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Merge branch 'master' into patch-1

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程序员Carl
2023-05-08 12:54:16 +08:00
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commit 37214e540f
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49
problems/0001.两数之和.md
View File

@ -151,7 +151,7 @@ public int[] twoSum(int[] nums, int target) {
```
Python：
（版本一） 使用字典
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
@ -163,6 +163,53 @@ class Solution:
records[value] = index # 遍历当前元素，并在map中寻找是否有匹配的key
return []
```
（版本二）使用集合
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
#创建一个集合来存储我们目前看到的数字
seen = set()
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
return [nums.index(complement), i]
seen.add(num)
```
（版本三）使用双指针
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
# 对输入列表进行排序
nums_sorted = sorted(nums)
# 使用双指针
left = 0
right = len(nums_sorted) - 1
while left < right:
current_sum = nums_sorted[left] + nums_sorted[right]
if current_sum == target:
# 如果和等于目标数，则返回两个数的下标
left_index = nums.index(nums_sorted[left])
right_index = nums.index(nums_sorted[right])
if left_index == right_index:
right_index = nums[left_index+1:].index(nums_sorted[right]) + left_index + 1
return [left_index, right_index]
elif current_sum < target:
# 如果总和小于目标，则将左侧指针向右移动
left += 1
else:
# 如果总和大于目标值，则将右指针向左移动
right -= 1
```
（版本四）暴力法
```python
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[i] + nums[j] == target:
return [i,j]
```
Go：

95
problems/0015.三数之和.md
View File

@ -298,61 +298,72 @@ class Solution {
```
Python：
（版本一） 双指针
```Python
class Solution:
def threeSum(self, nums):
ans = []
n = len(nums)
def threeSum(self, nums: List[int]) -> List[List[int]]:
result = []
nums.sort()
# 找出a + b + c = 0
# a = nums[i], b = nums[left], c = nums[right]
for i in range(n):
left = i + 1
right = n - 1
# 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
if nums[i] > 0:
break
if i >= 1 and nums[i] == nums[i - 1]: # 去重a
for i in range(len(nums)):
# 如果第一个元素已经大于0，不需要进一步检查
if nums[i] > 0:
return result
# 跳过相同的元素以避免重复
if i > 0 and nums[i] == nums[i - 1]:
continue
while left < right:
total = nums[i] + nums[left] + nums[right]
if total > 0:
right -= 1
elif total < 0:
left = i + 1
right = len(nums) - 1
while right > left:
sum_ = nums[i] + nums[left] + nums[right]
if sum_ < 0:
left += 1
elif sum_ > 0:
right -= 1
else:
ans.append([nums[i], nums[left], nums[right]])
# 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
while left != right and nums[left] == nums[left + 1]: left += 1
while left != right and nums[right] == nums[right - 1]: right -= 1
left += 1
result.append([nums[i], nums[left], nums[right]])
# 跳过相同的元素以避免重复
while right > left and nums[right] == nums[right - 1]:
right -= 1
while right > left and nums[left] == nums[left + 1]:
left += 1
right -= 1
return ans
left += 1
return result
```
Python (v3):
（版本二） 使用字典
```python
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
if len(nums) < 3: return []
nums, res = sorted(nums), []
for i in range(len(nums) - 2):
cur, l, r = nums[i], i + 1, len(nums) - 1
if res != [] and res[-1][0] == cur: continue # Drop duplicates for the first time.
while l < r:
if cur + nums[l] + nums[r] == 0:
res.append([cur, nums[l], nums[r]])
# Drop duplicates for the second time in interation of l & r. Only used when target situation occurs, because that is the reason for dropping duplicates.
while l < r - 1 and nums[l] == nums[l + 1]:
l += 1
while r > l + 1 and nums[r] == nums[r - 1]:
r -= 1
if cur + nums[l] + nums[r] > 0:
r -= 1
result = []
nums.sort()
# 找出a + b + c = 0
# a = nums[i], b = nums[j], c = -(a + b)
for i in range(len(nums)):
# 排序之后如果第一个元素已经大于零，那么不可能凑成三元组
if nums[i] > 0:
break
if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重
continue
d = {}
for j in range(i + 1, len(nums)):
if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重
continue
c = 0 - (nums[i] + nums[j])
if c in d:
result.append([nums[i], nums[j], c])
d.pop(c) # 三元组元素c去重
else:
l += 1
return res
d[nums[j]] = j
return result
```
Go：

84
problems/0018.四数之和.md
View File

@ -207,35 +207,44 @@ class Solution {
```
Python：
(版本一) 双指针
```python
# 双指针法
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
nums.sort()
n = len(nums)
res = []
for i in range(n):
if i > 0 and nums[i] == nums[i - 1]: continue # 对nums[i]去重
for k in range(i+1, n):
if k > i + 1 and nums[k] == nums[k-1]: continue # 对nums[k]去重
p = k + 1
q = n - 1
while p < q:
if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1
elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1
result = []
for i in range(n):
if nums[i] > target and nums[i] > 0 and target > 0:# 剪枝（可省）
break
if i > 0 and nums[i] == nums[i-1]:# 去重
continue
for j in range(i+1, n):
if nums[i] + nums[j] > target and target > 0: #剪枝（可省）
break
if j > i+1 and nums[j] == nums[j-1]: # 去重
continue
left, right = j+1, n-1
while left < right:
s = nums[i] + nums[j] + nums[left] + nums[right]
if s == target:
result.append([nums[i], nums[j], nums[left], nums[right]])
while left < right and nums[left] == nums[left+1]:
left += 1
while left < right and nums[right] == nums[right-1]:
right -= 1
left += 1
right -= 1
elif s < target:
left += 1
else:
res.append([nums[i], nums[k], nums[p], nums[q]])
# 对nums[p]和nums[q]去重
while p < q and nums[p] == nums[p + 1]: p += 1
while p < q and nums[q] == nums[q - 1]: q -= 1
p += 1
q -= 1
return res
right -= 1
return result
```
(版本二) 使用字典
```python
# 哈希表法
class Solution(object):
def fourSum(self, nums, target):
"""
@ -243,36 +252,25 @@ class Solution(object):
:type target: int
:rtype: List[List[int]]
"""
# use a dict to store value:showtimes
hashmap = dict()
for n in nums:
if n in hashmap:
hashmap[n] += 1
else:
hashmap[n] = 1
# 创建一个字典来存储输入列表中每个数字的频率
freq = {}
for num in nums:
freq[num] = freq.get(num, 0) + 1
# good thing about using python is you can use set to drop duplicates.
# 创建一个集合来存储最终答案，并遍历4个数字的所有唯一组合
ans = set()
# ans = [] # save results by list()
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
for k in range(j + 1, len(nums)):
val = target - (nums[i] + nums[j] + nums[k])
if val in hashmap:
# make sure no duplicates.
if val in freq:
# 确保没有重复
count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val)
if hashmap[val] > count:
ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val]))
ans.add(ans_tmp)
# Avoiding duplication in list manner but it cause time complexity increases
# if ans_tmp not in ans:
# ans.append(ans_tmp)
else:
continue
return list(ans)
# if used list() to save results, just
# return ans
if freq[val] > count:
ans.add(tuple(sorted([nums[i], nums[j], nums[k], val])))
return [list(x) for x in ans]
```
Go：

28
problems/0019.删除链表的倒数第N个节点.md
View File

@ -127,21 +127,29 @@ Python:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
head_dummy = ListNode()
head_dummy.next = head
slow, fast = head_dummy, head_dummy
while(n>=0): #fast先往前走n+1步
# 创建一个虚拟节点，并将其下一个指针设置为链表的头部
dummy_head = ListNode(0, head)
# 创建两个指针，慢指针和快指针，并将它们初始化为虚拟节点
slow = fast = dummy_head
# 快指针比慢指针快 n+1 步
for i in range(n+1):
fast = fast.next
n -= 1
while(fast!=None):
# 移动两个指针，直到快速指针到达链表的末尾
while fast:
slow = slow.next
fast = fast.next
#fast 走到结尾后，slow的下一个节点为倒数第N个节点
slow.next = slow.next.next #删除
return head_dummy.next
# 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点
slow.next = slow.next.next
return dummy_head.next
```
Go:
```Go

23
problems/0024.两两交换链表中的节点.md
View File

@ -186,21 +186,20 @@ Python：
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
res = ListNode(next=head)
pre = res
dummy_head = ListNode(next=head)
current = dummy_head
# 必须有pre的下一个和下下个才能交换，否则说明已经交换结束了
while pre.next and pre.next.next:
cur = pre.next
post = pre.next.next
# 必须有cur的下一个和下下个才能交换，否则说明已经交换结束了
while current.next and current.next.next:
temp = current.next # 防止节点修改
temp1 = current.next.next.next
# pre，cur，post对应最左，中间的，最右边的节点
cur.next = post.next
post.next = cur
pre.next = post
current.next = current.next.next
current.next.next = temp
temp.next = temp1
current = current.next.next
return dummy_head.next
pre = pre.next.next
return res.next
```
Go：

17
problems/0027.移除元素.md
View File

@ -198,6 +198,7 @@ Python：
``` python 3
（版本一）快慢指针法
class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
# 快慢指针
@ -213,7 +214,21 @@ class Solution:
return slow
```
``` python 3
（版本二）暴力法
class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
i, l = 0, len(nums)
while i < l:
if nums[i] == val: # 找到等于目标值的节点
for j in range(i+1, l): # 移除该元素，并将后面元素向前平移
nums[j - 1] = nums[j]
l -= 1
i -= 1
i += 1
return l
```
Go：

157
problems/0028.实现strStr.md
View File

@ -692,8 +692,67 @@ class Solution {
```
Python3：
（版本一）前缀表（减一）
```python
class Solution:
def getNext(self, next, s):
j = -1
next[0] = j
for i in range(1, len(s)):
while j >= 0 and s[i] != s[j+1]:
j = next[j]
if s[i] == s[j+1]:
j += 1
next[i] = j
def strStr(self, haystack: str, needle: str) -> int:
if not needle:
return 0
next = [0] * len(needle)
self.getNext(next, needle)
j = -1
for i in range(len(haystack)):
while j >= 0 and haystack[i] != needle[j+1]:
j = next[j]
if haystack[i] == needle[j+1]:
j += 1
if j == len(needle) - 1:
return i - len(needle) + 1
return -1
```
（版本二）前缀表（不减一）
```python
class Solution:
def getNext(self, next: List[int], s: str) -> None:
j = 0
next[0] = 0
for i in range(1, len(s)):
while j > 0 and s[i] != s[j]:
j = next[j - 1]
if s[i] == s[j]:
j += 1
next[i] = j
def strStr(self, haystack: str, needle: str) -> int:
if len(needle) == 0:
return 0
next = [0] * len(needle)
self.getNext(next, needle)
j = 0
for i in range(len(haystack)):
while j > 0 and haystack[i] != needle[j]:
j = next[j - 1]
if haystack[i] == needle[j]:
j += 1
if j == len(needle):
return i - len(needle) + 1
return -1
```
（版本三）暴力法
```python
//暴力解法：
class Solution(object):
def strStr(self, haystack, needle):
"""
@ -707,102 +766,22 @@ class Solution(object):
return i
return -1
```
（版本四）使用 index
```python
// 方法一
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
a = len(needle)
b = len(haystack)
if a == 0:
return 0
next = self.getnext(a,needle)
p=-1
for j in range(b):
while p >= 0 and needle[p+1] != haystack[j]:
p = next[p]
if needle[p+1] == haystack[j]:
p += 1
if p == a-1:
return j-a+1
return -1
def getnext(self,a,needle):
next = ['' for i in range(a)]
k = -1
next[0] = k
for i in range(1, len(needle)):
while (k > -1 and needle[k+1] != needle[i]):
k = next[k]
if needle[k+1] == needle[i]:
k += 1
next[i] = k
return next
```
```python
// 方法二
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
a = len(needle)
b = len(haystack)
if a == 0:
return 0
i = j = 0
next = self.getnext(a, needle)
while(i < b and j < a):
if j == -1 or needle[j] == haystack[i]:
i += 1
j += 1
else:
j = next[j]
if j == a:
return i-j
else:
try:
return haystack.index(needle)
except ValueError:
return -1
def getnext(self, a, needle):
next = ['' for i in range(a)]
j, k = 0, -1
next[0] = k
while(j < a-1):
if k == -1 or needle[k] == needle[j]:
k += 1
j += 1
next[j] = k
else:
k = next[k]
return next
```
（版本五）使用 find
```python
// 前缀表（不减一）Python实现
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
if len(needle) == 0:
return 0
next = self.getNext(needle)
j = 0
for i in range(len(haystack)):
while j >= 1 and haystack[i] != needle[j]:
j = next[j-1]
if haystack[i] == needle[j]:
j += 1
if j == len(needle):
return i - len(needle) + 1
return -1
def getNext(self, needle):
next = [0] * len(needle)
j = 0
next[0] = j
for i in range(1, len(needle)):
while j >= 1 and needle[i] != needle[j]:
j = next[j-1]
if needle[i] == needle[j]:
j += 1
next[i] = j
return next
```
return haystack.find(needle)
```
Go：

27
problems/0084.柱状图中最大的矩形.md
View File

@ -307,6 +307,33 @@ class Solution {
}
}
```
单调栈精简
```java
class Solution {
public int largestRectangleArea(int[] heights) {
int[] newHeight = new int[heights.length + 2];
System.arraycopy(heights, 0, newHeight, 1, heights.length);
newHeight[heights.length+1] = 0;
newHeight[0] = 0;
Stack<Integer> stack = new Stack<>();
stack.push(0);
int res = 0;
for (int i = 1; i < newHeight.length; i++) {
while (newHeight[i] < newHeight[stack.peek()]) {
int mid = stack.pop();
int w = i - stack.peek() - 1;
int h = newHeight[mid];
res = Math.max(res, w * h);
}
stack.push(i);
}
return res;
}
}
```
Python3:

40
problems/0101.对称二叉树.md
View File

@ -442,25 +442,31 @@ class Solution:
层次遍历
```python
class Solution:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
que = [root]
while que:
this_level_length = len(que)
for i in range(this_level_length // 2):
# 要么其中一个是None但另外一个不是
if (not que[i] and que[this_level_length - 1 - i]) or (que[i] and not que[this_level_length - 1 - i]):
return False
# 要么两个都不是None
if que[i] and que[i].val != que[this_level_length - 1 - i].val:
return False
for i in range(this_level_length):
if not que[i]: continue
que.append(que[i].left)
que.append(que[i].right)
que = que[this_level_length:]
queue = collections.deque([root.left, root.right])
while queue:
level_size = len(queue)
if level_size % 2 != 0:
return False
level_vals = []
for i in range(level_size):
node = queue.popleft()
if node:
level_vals.append(node.val)
queue.append(node.left)
queue.append(node.right)
else:
level_vals.append(None)
if level_vals != level_vals[::-1]:
return False
return True
```

438
problems/0102.二叉树的层序遍历.md
View File

@ -171,47 +171,59 @@ python3代码：
```python
# 利用长度法
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
"""二叉树层序遍历迭代解法"""
def levelOrder(self, root: TreeNode) -> List[List[int]]:
results = []
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
if not root:
return results
from collections import deque
que = deque([root])
while que:
size = len(que)
result = []
for _ in range(size):
cur = que.popleft()
result.append(cur.val)
return []
queue = collections.deque([root])
result = []
while queue:
level = []
for _ in range(len(queue)):
cur = queue.popleft()
level.append(cur.val)
if cur.left:
que.append(cur.left)
queue.append(cur.left)
if cur.right:
que.append(cur.right)
results.append(result)
return results
queue.append(cur.right)
result.append(level)
return result
```
```python
# 递归法
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
res = []
def helper(root, depth):
if not root: return []
if len(res) == depth: res.append([]) # start the current depth
res[depth].append(root.val) # fulfil the current depth
if root.left: helper(root.left, depth + 1) # process child nodes for the next depth
if root.right: helper(root.right, depth + 1)
helper(root, 0)
return res
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
levels = []
self.helper(root, 0, levels)
return levels
def helper(self, node, level, levels):
if not node:
return
if len(levels) == level:
levels.append([])
levels[level].append(node.val)
self.helper(node.left, level + 1, levels)
self.helper(node.right, level + 1, levels)
```
go:
```go
@ -500,27 +512,29 @@ python代码：
class Solution:
"""二叉树层序遍历II迭代解法"""
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
results = []
if not root:
return results
from collections import deque
que = deque([root])
while que:
result = []
for _ in range(len(que)):
cur = que.popleft()
result.append(cur.val)
return []
queue = collections.deque([root])
result = []
while queue:
level = []
for _ in range(len(queue)):
cur = queue.popleft()
level.append(cur.val)
if cur.left:
que.append(cur.left)
queue.append(cur.left)
if cur.right:
que.append(cur.right)
results.append(result)
results.reverse()
return results
queue.append(cur.right)
result.append(level)
return result[::-1]
```
Java：
@ -821,35 +835,35 @@ public:
python代码：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root:
return []
# deque来自collections模块，不在力扣平台时，需要手动写入
# 'from collections import deque' 导入
# deque相比list的好处是，list的pop(0)是O(n)复杂度，deque的popleft()是O(1)复杂度
quene = deque([root])
out_list = []
while quene:
# 每次都取最后一个node就可以了
node = quene[-1]
out_list.append(node.val)
# 执行这个遍历的目的是获取下一层所有的node
for _ in range(len(quene)):
node = quene.popleft()
queue = collections.deque([root])
right_view = []
while queue:
level_size = len(queue)
for i in range(level_size):
node = queue.popleft()
if i == level_size - 1:
right_view.append(node.val)
if node.left:
quene.append(node.left)
queue.append(node.left)
if node.right:
quene.append(node.right)
return out_list
# 执行用时：36 ms, 在所有 Python3 提交中击败了89.47%的用户
# 内存消耗：14.6 MB, 在所有 Python3 提交中击败了96.65%的用户
queue.append(node.right)
return right_view
```
@ -1107,27 +1121,38 @@ python代码：
class Solution:
"""二叉树层平均值迭代解法"""
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def averageOfLevels(self, root: TreeNode) -> List[float]:
results = []
if not root:
return results
return []
from collections import deque
que = deque([root])
while que:
size = len(que)
sum_ = 0
for _ in range(size):
cur = que.popleft()
sum_ += cur.val
if cur.left:
que.append(cur.left)
if cur.right:
que.append(cur.right)
results.append(sum_ / size)
return results
queue = collections.deque([root])
averages = []
while queue:
size = len(queue)
level_sum = 0
for i in range(size):
node = queue.popleft()
level_sum += node.val
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
averages.append(level_sum / size)
return averages
```
java:
@ -1401,28 +1426,36 @@ public:
python代码：
```python
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
"""N叉树的层序遍历迭代法"""
def levelOrder(self, root: 'Node') -> List[List[int]]:
results = []
if not root:
return results
return []
from collections import deque
que = deque([root])
result = []
queue = collections.deque([root])
while que:
result = []
for _ in range(len(que)):
cur = que.popleft()
result.append(cur.val)
# cur.children 是 Node 对象组成的列表，也可能为 None
if cur.children:
que.extend(cur.children)
results.append(result)
while queue:
level_size = len(queue)
level = []
return results
for _ in range(level_size):
node = queue.popleft()
level.append(node.val)
for child in node.children:
queue.append(child)
result.append(level)
return result
```
```python
@ -1728,22 +1761,37 @@ public:
python代码：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def largestValues(self, root: TreeNode) -> List[int]:
if root is None:
if not root:
return []
queue = [root]
out_list = []
result = []
queue = collections.deque([root])
while queue:
length = len(queue)
in_list = []
for _ in range(length):
curnode = queue.pop(0)
in_list.append(curnode.val)
if curnode.left: queue.append(curnode.left)
if curnode.right: queue.append(curnode.right)
out_list.append(max(in_list))
return out_list
level_size = len(queue)
max_val = float('-inf')
for _ in range(level_size):
node = queue.popleft()
max_val = max(max_val, node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(max_val)
return result
```
java代码：
@ -2048,36 +2096,40 @@ class Solution {
python代码：
```python
# 层序遍历解法
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
queue = [root]
return root
queue = collections.deque([root])
while queue:
n = len(queue)
for i in range(n):
node = queue.pop(0)
level_size = len(queue)
prev = None
for i in range(level_size):
node = queue.popleft()
if prev:
prev.next = node
prev = node
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if i == n - 1:
break
node.next = queue[0]
return root
# 链表解法
class Solution:
def connect(self, root: 'Node') -> 'Node':
first = root
while first:
cur = first
while cur: # 遍历每一层的节点
if cur.left: cur.left.next = cur.right # 找左节点的next
if cur.right and cur.next: cur.right.next = cur.next.left # 找右节点的next
cur = cur.next # cur同层移动到下一节点
first = first.left # 从本层扩展到下一层
return root
```
@ -2329,21 +2381,41 @@ python代码：
```python
# 层序遍历解法
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
queue = [root]
while queue: # 遍历每一层
length = len(queue)
tail = None # 每一层维护一个尾节点
for i in range(length): # 遍历当前层
curnode = queue.pop(0)
if tail:
tail.next = curnode # 让尾节点指向当前节点
tail = curnode # 让当前节点成为尾节点
if curnode.left : queue.append(curnode.left)
if curnode.right: queue.append(curnode.right)
return root
queue = collections.deque([root])
while queue:
level_size = len(queue)
prev = None
for i in range(level_size):
node = queue.popleft()
if prev:
prev.next = node
prev = node
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return root
```
@ -2592,24 +2664,31 @@ class Solution {
Python：
```python 3
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def maxDepth(self, root: TreeNode) -> int:
if root == None:
if not root:
return 0
queue_ = [root]
depth = 0
while queue_:
length = len(queue_)
for i in range(length):
cur = queue_.pop(0)
sub.append(cur.val)
#子节点入队列
if cur.left: queue_.append(cur.left)
if cur.right: queue_.append(cur.right)
queue = collections.deque([root])
while queue:
depth += 1
for _ in range(len(queue)):
node = queue.popleft()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return depth
```
Go：
@ -2859,23 +2938,26 @@ Python 3：
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if root == None:
if not root:
return 0
depth = 0
queue = collections.deque([root])
while queue:
depth += 1
for _ in range(len(queue)):
node = queue.popleft()
if not node.left and not node.right:
return depth
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
#根节点的深度为1
queue_ = [(root,1)]
while queue_:
cur, depth = queue_.pop(0)
if cur.left == None and cur.right == None:
return depth
#先左子节点，由于左子节点没有孩子，则就是这一层了
if cur.left:
queue_.append((cur.left,depth + 1))
if cur.right:
queue_.append((cur.right,depth + 1))
return 0
return depth
```
Go：

121
problems/0104.二叉树的最大深度.md
View File

@ -419,86 +419,107 @@ class solution:
return 1 + max(self.maxdepth(root.left), self.maxdepth(root.right))
```
迭代法：
层序遍历迭代法：
```python
import collections
class solution:
def maxdepth(self, root: treenode) -> int:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def maxDepth(self, root: TreeNode) -> int:
if not root:
return 0
depth = 0 #记录深度
queue = collections.deque()
queue.append(root)
depth = 0
queue = collections.deque([root])
while queue:
size = len(queue)
depth += 1
for i in range(size):
for _ in range(len(queue)):
node = queue.popleft()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return depth
```
### 559.n叉树的最大深度
递归法：
```python
class solution:
def maxdepth(self, root: 'node') -> int:
class Solution:
def maxDepth(self, root: 'Node') -> int:
if not root:
return 0
depth = 0
for i in range(len(root.children)):
depth = max(depth, self.maxdepth(root.children[i]))
return depth + 1
max_depth = 1
for child in root.children:
max_depth = max(max_depth, self.maxDepth(child) + 1)
return max_depth
```
迭代法：
```python
import collections
class solution:
def maxdepth(self, root: 'node') -> int:
queue = collections.deque()
if root:
queue.append(root)
depth = 0 #记录深度
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def maxDepth(self, root: TreeNode) -> int:
if not root:
return 0
depth = 0
queue = collections.deque([root])
while queue:
size = len(queue)
depth += 1
for i in range(size):
for _ in range(len(queue)):
node = queue.popleft()
for j in range(len(node.children)):
if node.children[j]:
queue.append(node.children[j])
for child in node.children:
queue.append(child)
return depth
```
使用栈来模拟后序遍历依然可以
使用栈
```python
class solution:
def maxdepth(self, root: 'node') -> int:
st = []
if root:
st.append(root)
depth = 0
result = 0
while st:
node = st.pop()
if node != none:
st.append(node) #中
st.append(none)
depth += 1
for i in range(len(node.children)): #处理孩子
if node.children[i]:
st.append(node.children[i])
else:
node = st.pop()
depth -= 1
result = max(result, depth)
return result
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def maxDepth(self, root: 'Node') -> int:
if not root:
return 0
max_depth = 0
stack = [(root, 1)]
while stack:
node, depth = stack.pop()
max_depth = max(max_depth, depth)
for child in node.children:
stack.append((child, depth + 1))
return max_depth
```

35
problems/0106.从中序与后序遍历序列构造二叉树.md
View File

@ -622,7 +622,42 @@ class Solution {
}
}
```
```java
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if(postorder.length == 0 || inorder.length == 0)
return null;
return buildHelper(inorder, 0, inorder.length, postorder, 0, postorder.length);
}
private TreeNode buildHelper(int[] inorder, int inorderStart, int inorderEnd, int[] postorder, int postorderStart, int postorderEnd){
if(postorderStart == postorderEnd)
return null;
int rootVal = postorder[postorderEnd - 1];
TreeNode root = new TreeNode(rootVal);
int middleIndex;
for (middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++){
if(inorder[middleIndex] == rootVal)
break;
}
int leftInorderStart = inorderStart;
int leftInorderEnd = middleIndex;
int rightInorderStart = middleIndex + 1;
int rightInorderEnd = inorderEnd;
int leftPostorderStart = postorderStart;
int leftPostorderEnd = postorderStart + (middleIndex - inorderStart);
int rightPostorderStart = leftPostorderEnd;
int rightPostorderEnd = postorderEnd - 1;
root.left = buildHelper(inorder, leftInorderStart, leftInorderEnd, postorder, leftPostorderStart, leftPostorderEnd);
root.right = buildHelper(inorder, rightInorderStart, rightInorderEnd, postorder, rightPostorderStart, rightPostorderEnd);
return root;
}
}
```
105.从前序与中序遍历序列构造二叉树
```java

18
problems/0110.平衡二叉树.md
View File

@ -532,6 +532,24 @@ class Solution:
else:
return 1 + max(left_height, right_height)
```
递归法精简版：
```python
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
return self.height(root) != -1
def height(self, node: TreeNode) -> int:
if not node:
return 0
left = self.height(node.left)
if left == -1:
return -1
right = self.height(node.right)
if right == -1 or abs(left - right) > 1:
return -1
return max(left, right) + 1
```
迭代法：

92
problems/0111.二叉树的最小深度.md
View File

@ -305,46 +305,98 @@ class Solution {
递归法：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
if not root.left and not root.right:
return 1
min_depth = 10**9
left_depth = float('inf')
right_depth = float('inf')
if root.left:
min_depth = min(self.minDepth(root.left), min_depth) # 获得左子树的最小高度
left_depth = self.minDepth(root.left)
if root.right:
min_depth = min(self.minDepth(root.right), min_depth) # 获得右子树的最小高度
return min_depth + 1
right_depth = self.minDepth(root.right)
return 1 + min(left_depth, right_depth)
```
迭代法：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
que = deque()
que.append(root)
res = 1
while que:
for _ in range(len(que)):
node = que.popleft()
# 当左右孩子都为空的时候，说明是最低点的一层了，退出
depth = 0
queue = collections.deque([root])
while queue:
depth += 1
for _ in range(len(queue)):
node = queue.popleft()
if not node.left and not node.right:
return res
if node.left is not None:
que.append(node.left)
if node.right is not None:
que.append(node.right)
res += 1
return res
return depth
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return depth
```
迭代法：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
queue = collections.deque([(root, 1)])
while queue:
node, depth = queue.popleft()
# Check if the node is a leaf node
if not node.left and not node.right:
return depth
# Add left and right child to the queue
if node.left:
queue.append((node.left, depth+1))
if node.right:
queue.append((node.right, depth+1))
return 0
```
## Go

267
problems/0112.路径总和.md
View File

@ -487,140 +487,191 @@ class Solution {
### 0112.路径总和
**递归**
(版本一) 递归
```python
class solution:
def haspathsum(self, root: treenode, targetsum: int) -> bool:
def isornot(root, targetsum) -> bool:
if (not root.left) and (not root.right) and targetsum == 0:
return true # 遇到叶子节点，并且计数为0
if (not root.left) and (not root.right):
return false # 遇到叶子节点，计数不为0
if root.left:
targetsum -= root.left.val # 左节点
if isornot(root.left, targetsum): return true # 递归，处理左节点
targetsum += root.left.val # 回溯
if root.right:
targetsum -= root.right.val # 右节点
if isornot(root.right, targetsum): return true # 递归，处理右节点
targetsum += root.right.val # 回溯
return false
if root == none:
return false # 别忘记处理空treenode
else:
return isornot(root, targetsum - root.val)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def traversal(self, cur: TreeNode, count: int) -> bool:
if not cur.left and not cur.right and count == 0: # 遇到叶子节点，并且计数为0
return True
if not cur.left and not cur.right: # 遇到叶子节点直接返回
return False
if cur.left: # 左
count -= cur.left.val
if self.traversal(cur.left, count): # 递归，处理节点
return True
count += cur.left.val # 回溯，撤销处理结果
class Solution: # 简洁版
def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
if not root: return False
if root.left==root.right==None and root.val == targetSum: return True
return self.hasPathSum(root.left,targetSum-root.val) or self.hasPathSum(root.right,targetSum-root.val)
if cur.right: # 右
count -= cur.right.val
if self.traversal(cur.right, count): # 递归，处理节点
return True
count += cur.right.val # 回溯，撤销处理结果
return False
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if root is None:
return False
return self.traversal(root, sum - root.val)
```
**迭代 - 层序遍历**
(版本二) 递归 + 精简
```python
class solution:
def haspathsum(self, root: treenode, targetsum: int) -> bool:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if not root:
return false
stack = [] # [(当前节点，路径数值), ...]
stack.append((root, root.val))
while stack:
cur_node, path_sum = stack.pop()
if not cur_node.left and not cur_node.right and path_sum == targetsum:
return true
if cur_node.right:
stack.append((cur_node.right, path_sum + cur_node.right.val))
if cur_node.left:
stack.append((cur_node.left, path_sum + cur_node.left.val))
return false
return False
if not root.left and not root.right and sum == root.val:
return True
return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
```
(版本三) 迭代
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if not root:
return False
# 此时栈里要放的是pair<节点指针，路径数值>
st = [(root, root.val)]
while st:
node, path_sum = st.pop()
# 如果该节点是叶子节点了，同时该节点的路径数值等于sum，那么就返回true
if not node.left and not node.right and path_sum == sum:
return True
# 右节点，压进去一个节点的时候，将该节点的路径数值也记录下来
if node.right:
st.append((node.right, path_sum + node.right.val))
# 左节点，压进去一个节点的时候，将该节点的路径数值也记录下来
if node.left:
st.append((node.left, path_sum + node.left.val))
return False
```
### 0113.路径总和-ii
**递归**
(版本一) 递归
```python
class solution:
def pathsum(self, root: treenode, targetsum: int) -> list[list[int]]:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def __init__(self):
self.result = []
self.path = []
def traversal(cur_node, remain):
if not cur_node.left and not cur_node.right:
if remain == 0:
result.append(path[:])
return
def traversal(self, cur, count):
if not cur.left and not cur.right and count == 0: # 遇到了叶子节点且找到了和为sum的路径
self.result.append(self.path[:])
return
if cur_node.left:
path.append(cur_node.left.val)
traversal(cur_node.left, remain-cur_node.left.val)
path.pop()
if not cur.left and not cur.right: # 遇到叶子节点而没有找到合适的边，直接返回
return
if cur_node.right:
path.append(cur_node.right.val)
traversal(cur_node.right, remain-cur_node.right.val)
path.pop()
if cur.left: # 左 （空节点不遍历）
self.path.append(cur.left.val)
count -= cur.left.val
self.traversal(cur.left, count) # 递归
count += cur.left.val # 回溯
self.path.pop() # 回溯
result, path = [], []
if cur.right: # 右 （空节点不遍历）
self.path.append(cur.right.val)
count -= cur.right.val
self.traversal(cur.right, count) # 递归
count += cur.right.val # 回溯
self.path.pop() # 回溯
return
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
self.result.clear()
self.path.clear()
if not root:
return []
path.append(root.val)
traversal(root, targetsum - root.val)
return result
return self.result
self.path.append(root.val) # 把根节点放进路径
self.traversal(root, sum - root.val)
return self.result
```
**迭代法，用第二个队列保存目前的总和与路径**
(版本二) 递归 + 精简
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
if not root:
return []
que, temp = deque([root]), deque([(root.val, [root.val])])
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
result = []
while que:
for _ in range(len(que)):
node = que.popleft()
value, path = temp.popleft()
if (not node.left) and (not node.right):
if value == targetSum:
result.append(path)
if node.left:
que.append(node.left)
temp.append((node.left.val+value, path+[node.left.val]))
if node.right:
que.append(node.right)
temp.append((node.right.val+value, path+[node.right.val]))
self.traversal(root, targetSum, [], result)
return result
def traversal(self,node, count, path, result):
if not node:
return
path.append(node.val)
count -= node.val
if not node.left and not node.right and count == 0:
result.append(list(path))
self.traversal(node.left, count, path, result)
self.traversal(node.right, count, path, result)
path.pop()
```
**迭代法，前序遍历**
(版本三) 迭代
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
if not root: return []
stack, path_stack,result = [[root,root.val]],[[root.val]],[]
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
if not root:
return []
stack = [(root, [root.val])]
res = []
while stack:
cur,cursum = stack.pop()
path = path_stack.pop()
if cur.left==cur.right==None:
if cursum==targetSum: result.append(path)
if cur.right:
stack.append([cur.right,cursum+cur.right.val])
path_stack.append(path+[cur.right.val])
if cur.left:
stack.append([cur.left,cursum+cur.left.val])
path_stack.append(path+[cur.left.val])
return result
node, path = stack.pop()
if not node.left and not node.right and sum(path) == targetSum:
res.append(path)
if node.right:
stack.append((node.right, path + [node.right.val]))
if node.left:
stack.append((node.left, path + [node.left.val]))
return res
```
## go

50
problems/0142.环形链表II.md
View File

@ -221,25 +221,55 @@ public class Solution {
Python：
```python
（版本一）快慢指针法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def detectCycle(self, head: ListNode) -> ListNode:
slow, fast = head, head
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
# 如果相遇
# If there is a cycle, the slow and fast pointers will eventually meet
if slow == fast:
p = head
q = slow
while p!=q:
p = p.next
q = q.next
#你也可以return q
return p
# Move one of the pointers back to the start of the list
slow = head
while slow != fast:
slow = slow.next
fast = fast.next
return slow
# If there is no cycle, return None
return None
```
```python
（版本二）集合法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def detectCycle(self, head: ListNode) -> ListNode:
visited = set()
while head:
if head in visited:
return head
visited.add(head)
head = head.next
return None
```
Go：
```go

136
problems/0151.翻转字符串里的单词.md
View File

@ -434,134 +434,40 @@ class Solution {
```
python:
（版本一）先删除空白，然后整个反转，最后单词反转。
**因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1)**
```Python
class Solution:
#1.去除多余的空格
def trim_spaces(self, s):
n = len(s)
left = 0
right = n-1
while left <= right and s[left] == ' ': #去除开头的空格
left += 1
while left <= right and s[right] == ' ': #去除结尾的空格
right -= 1
tmp = []
while left <= right: #去除单词中间多余的空格
if s[left] != ' ':
tmp.append(s[left])
elif tmp[-1] != ' ': #当前位置是空格，但是相邻的上一个位置不是空格，则该空格是合理的
tmp.append(s[left])
left += 1
return tmp
#2.翻转字符数组
def reverse_string(self, nums, left, right):
while left < right:
nums[left], nums[right] = nums[right], nums[left]
left += 1
right -= 1
return None
#3.翻转每个单词
def reverse_each_word(self, nums):
start = 0
end = 0
n = len(nums)
while start < n:
while end < n and nums[end] != ' ':
end += 1
self.reverse_string(nums, start, end-1)
start = end + 1
end += 1
return None
#4.翻转字符串里的单词
def reverseWords(self, s): #测试用例："the sky is blue"
l = self.trim_spaces(s) #输出：['t', 'h', 'e', ' ', 's', 'k', 'y', ' ', 'i', 's', ' ', 'b', 'l', 'u', 'e'
self.reverse_string(l, 0, len(l)-1) #输出：['e', 'u', 'l', 'b', ' ', 's', 'i', ' ', 'y', 'k', 's', ' ', 'e', 'h', 't']
self.reverse_each_word(l) #输出：['b', 'l', 'u', 'e', ' ', 'i', 's', ' ', 's', 'k', 'y', ' ', 't', 'h', 'e']
return ''.join(l) #输出：blue is sky the
def reverseWords(self, s: str) -> str:
# 删除前后空白
s = s.strip()
# 反转整个字符串
s = s[::-1]
# 将字符串拆分为单词，并反转每个单词
s = ' '.join(word[::-1] for word in s.split())
return s
```
（版本二）使用双指针
```python
class Solution:
def reverseWords(self, s: str) -> str:
# method 1 - Rude but work & efficient method.
s_list = [i for i in s.split(" ") if len(i) > 0]
return " ".join(s_list[::-1])
# 将字符串拆分为单词，即转换成列表类型
words = s.split()
# method 2 - Carlo's idea
def trim_head_tail_space(ss: str):
p = 0
while p < len(ss) and ss[p] == " ":
p += 1
return ss[p:]
# 反转单词
left, right = 0, len(words) - 1
while left < right:
words[left], words[right] = words[right], words[left]
left += 1
right -= 1
# Trim the head and tail space
s = trim_head_tail_space(s)
s = trim_head_tail_space(s[::-1])[::-1]
pf, ps, s = 0, 0, s[::-1] # Reverse the string.
while pf < len(s):
if s[pf] == " ":
# Will not excede. Because we have clean the tail space.
if s[pf] == s[pf + 1]:
s = s[:pf] + s[pf + 1:]
continue
else:
s = s[:ps] + s[ps: pf][::-1] + s[pf:]
ps, pf = pf + 1, pf + 2
else:
pf += 1
return s[:ps] + s[ps:][::-1] # Must do the last step, because the last word is omit though the pointers are on the correct positions,
# 将列表转换成字符串
return " ".join(words)
```
```python
class Solution: # 使用双指针法移除空格
def reverseWords(self, s: str) -> str:
def removeextraspace(s):
start = 0; end = len(s)-1
while s[start]==' ':
start+=1
while s[end]==' ':
end-=1
news = list(s[start:end+1])
slow = fast = 0
while fast<len(news):
while fast>0 and news[fast]==news[fast-1]==' ':
fast+=1
news[slow]=news[fast]
slow+=1; fast+=1
#return "".join(news[:slow])
return news[:slow]
def reversestr(s):
left,right = 0,len(s)-1
news = list(s)
while left<right:
news[left],news[right] = news[right],news[left]
left+=1; right-=1
#return "".join(news)
return news
news = removeextraspace(s)
news.append(' ')
fast=slow=0
#print(news)
while fast<len(news):
while news[fast]!=' ':
fast+=1
news[slow:fast] = reversestr(news[slow:fast])
# print(news[slow:fast])
fast=slow=fast+1
news2 = reversestr(news[:-1])
return ''.join(news2)
```
Go：
```go

96
problems/0202.快乐数.md
View File

@ -108,23 +108,14 @@ class Solution {
```
Python：
(版本一)使用集合
```python
class Solution:
def isHappy(self, n: int) -> bool:
def calculate_happy(num):
sum_ = 0
# 从个位开始依次取，平方求和
while num:
sum_ += (num % 10) ** 2
num = num // 10
return sum_
# 记录中间结果
def isHappy(self, n: int) -> bool:
record = set()
while True:
n = calculate_happy(n)
n = self.get_sum(n)
if n == 1:
return True
@ -134,21 +125,86 @@ class Solution:
else:
record.add(n)
# python的另一种写法 - 通过字符串来计算各位平方和
def get_sum(self,n: int) -> int:
new_num = 0
while n:
n, r = divmod(n, 10)
new_num += r ** 2
return new_num
```
(版本二)使用集合
```python
class Solution:
def isHappy(self, n: int) -> bool:
record = set()
while n not in record:
record.add(n)
new_num = 0
n_str = str(n)
for i in n_str:
new_num+=int(i)**2
if new_num==1: return True
else: n = new_num
return False
```
(版本三)使用数组
```python
class Solution:
def isHappy(self, n: int) -> bool:
record = []
while n not in record:
record.append(n)
newn = 0
nn = str(n)
for i in nn:
newn+=int(i)**2
if newn==1: return True
n = newn
new_num = 0
n_str = str(n)
for i in n_str:
new_num+=int(i)**2
if new_num==1: return True
else: n = new_num
return False
```
(版本四)使用快慢指针
```python
class Solution:
def isHappy(self, n: int) -> bool:
slow = n
fast = n
while self.get_sum(fast) != 1 and self.get_sum(self.get_sum(fast)):
slow = self.get_sum(slow)
fast = self.get_sum(self.get_sum(fast))
if slow == fast:
return False
return True
def get_sum(self,n: int) -> int:
new_num = 0
while n:
n, r = divmod(n, 10)
new_num += r ** 2
return new_num
```
(版本五)使用集合+精简
```python
class Solution:
def isHappy(self, n: int) -> bool:
seen = set()
while n != 1:
n = sum(int(i) ** 2 for i in str(n))
if n in seen:
return False
seen.add(n)
return True
```
(版本六)使用数组+精简
```python
class Solution:
def isHappy(self, n: int) -> bool:
seen = []
while n != 1:
n = sum(int(i) ** 2 for i in str(n))
if n in seen:
return False
seen.append(n)
return True
```
Go：
```go
func isHappy(n int) bool {

20
problems/0203.移除链表元素.md
View File

@ -307,21 +307,27 @@ public ListNode removeElements(ListNode head, int val) {
Python：
```python
（版本一）虚拟头节点法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeElements(self, head: ListNode, val: int) -> ListNode:
dummy_head = ListNode(next=head) #添加一个虚拟节点
cur = dummy_head
while cur.next:
if cur.next.val == val:
cur.next = cur.next.next #删除cur.next节点
def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
# 创建虚拟头部节点以简化删除过程
dummy_head = ListNode(next = head)
# 遍历列表并删除值为val的节点
current = dummy_head
while current.next:
if current.next.val == val:
current.next = current.next.next
else:
cur = cur.next
current = current.next
return dummy_head.next
```
Go：

40
problems/0206.翻转链表.md
View File

@ -193,9 +193,9 @@ class Solution {
}
```
Python迭代法：
Python
```python
#双指针
（版本一）双指针法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
@ -205,7 +205,7 @@ class Solution:
def reverseList(self, head: ListNode) -> ListNode:
cur = head
pre = None
while(cur!=None):
while cur:
temp = cur.next # 保存一下 cur的下一个节点，因为接下来要改变cur->next
cur.next = pre #反转
#更新pre、cur指针
@ -217,6 +217,7 @@ class Solution:
Python递归法：
```python
（版本二）递归法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
@ -224,36 +225,17 @@ Python递归法：
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
def reverse(pre,cur):
if not cur:
return pre
tmp = cur.next
cur.next = pre
return reverse(cur,tmp)
return reverse(None,head)
return self.reverse(head, None)
def reverse(self, cur: ListNode, pre: ListNode) -> ListNode:
if cur == None:
return pre
temp = cur.next
cur.next = pre
return self.reverse(temp, cur)
```
Python递归法从后向前：
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head or not head.next: return head
p = self.reverseList(head.next)
head.next.next = head
head.next = None
return p
```
Go：

46
problems/0209.长度最小的子数组.md
View File

@ -173,18 +173,44 @@ class Solution {
Python：
```python
（版本一）滑动窗口法
class Solution:
def minSubArrayLen(self, s: int, nums: List[int]) -> int:
res = float("inf") # 定义一个无限大的数
Sum = 0 # 滑动窗口数值之和
i = 0 # 滑动窗口起始位置
for j in range(len(nums)):
Sum += nums[j]
while Sum >= s:
res = min(res, j-i+1)
Sum -= nums[i]
i += 1
return 0 if res == float("inf") else res
l = len(nums)
left = 0
right = 0
min_len = float('inf')
cur_sum = 0 #当前的累加值
while right < l:
cur_sum += nums[right]
while cur_sum >= s: # 当前累加值大于目标值
min_len = min(min_len, right - left + 1)
cur_sum -= nums[left]
left += 1
right += 1
return min_len if min_len != float('inf') else 0
```
```python
（版本二）暴力法
class Solution:
def minSubArrayLen(self, s: int, nums: List[int]) -> int:
l = len(nums)
min_len = float('inf')
for i in range(l):
cur_sum = 0
for j in range(i, l):
cur_sum += nums[j]
if cur_sum >= s:
min_len = min(min_len, j - i + 1)
break
return min_len if min_len != float('inf') else 0
```
Go：

14
problems/0222.完全二叉树的节点个数.md
View File

@ -393,6 +393,20 @@ class Solution: # 利用完全二叉树特性
return 2**count-1
return 1+self.countNodes(root.left)+self.countNodes(root.right)
```
完全二叉树写法3
```python
class Solution: # 利用完全二叉树特性
def countNodes(self, root: TreeNode) -> int:
if not root: return 0
count = 0
left = root.left; right = root.right
while left and right:
count+=1
left = left.left; right = right.right
if not left and not right: # 如果同时到底说明是满二叉树，反之则不是
return (2<<count)-1
return 1+self.countNodes(root.left)+self.countNodes(root.right)
```
## Go

163
problems/0226.翻转二叉树.md
View File

@ -314,81 +314,158 @@ class Solution {
递归法：前序遍历：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if not root:
return None
root.left, root.right = root.right, root.left #中
self.invertTree(root.left) #左
self.invertTree(root.right) #右
root.left, root.right = root.right, root.left
self.invertTree(root.left)
self.invertTree(root.right)
return root
```
递归法：后序遍历：
迭代法：前序遍历：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if root is None:
if not root:
return None
stack = [root]
while stack:
node = stack.pop()
node.left, node.right = node.right, node.left
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return root
```
递归法：中序遍历：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if not root:
return None
self.invertTree(root.left)
root.left, root.right = root.right, root.left
self.invertTree(root.left)
return root
```
迭代法：中序遍历：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if not root:
return None
stack = [root]
while stack:
node = stack.pop()
if node.left:
stack.append(node.left)
node.left, node.right = node.right, node.left
if node.left:
stack.append(node.left)
return root
```
递归法：后序遍历：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if not root:
return None
self.invertTree(root.left)
self.invertTree(root.right)
root.left, root.right = root.right, root.left
return root
return root
```
迭代法：深度优先遍历（前序遍历）：
迭代法：后序遍历：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if not root:
return root
st = []
st.append(root)
while st:
node = st.pop()
node.left, node.right = node.right, node.left #中
if node.right:
st.append(node.right) #右
return None
stack = [root]
while stack:
node = stack.pop()
if node.left:
st.append(node.left) #左
stack.append(node.left)
if node.right:
stack.append(node.right)
node.left, node.right = node.right, node.left
return root
```
迭代法：广度优先遍历（层序遍历）：
```python
import collections
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
queue = collections.deque() #使用deque()
if root:
queue.append(root)
if not root:
return None
queue = collections.deque([root])
while queue:
size = len(queue)
for i in range(size):
for i in range(len(queue)):
node = queue.popleft()
node.left, node.right = node.right, node.left #节点处理
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return root
```
迭代法：广度优先遍历（层序遍历），和之前的层序遍历写法一致：
```python
class Solution:
def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if not root: return root
from collections import deque
que=deque([root])
while que:
size=len(que)
for i in range(size):
cur=que.popleft()
cur.left, cur.right = cur.right, cur.left
if cur.left: que.append(cur.left)
if cur.right: que.append(cur.right)
node.left, node.right = node.right, node.left
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
return root
```
### Go
递归版本的前序遍历

5
problems/0242.有效的字母异位词.md
View File

@ -122,6 +122,7 @@ class Solution {
```
Python：
```python
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
@ -137,7 +138,6 @@ class Solution:
return False
return True
```
Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）：
```python
@ -147,13 +147,11 @@ class Solution:
s_dict = defaultdict(int)
t_dict = defaultdict(int)
for x in s:
s_dict[x] += 1
for x in t:
t_dict[x] += 1
return s_dict == t_dict
```
Python写法三(没有使用数组作为哈希表，只是介绍Counter这种更方便的解题思路)：
@ -165,7 +163,6 @@ class Solution(object):
a_count = Counter(s)
b_count = Counter(t)
return a_count == b_count
```
Go：

65
problems/0257.二叉树的所有路径.md
View File

@ -468,6 +468,71 @@ class Solution {
```
---
## Python:
递归法+回溯（版本一）
```Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
import copy
from typing import List, Optional
class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
if not root:
return []
result = []
self.generate_paths(root, [], result)
return result
def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None:
path.append(node.val)
if not node.left and not node.right:
result.append('->'.join(map(str, path)))
if node.left:
self.generate_paths(node.left, copy.copy(path), result)
if node.right:
self.generate_paths(node.right, copy.copy(path), result)
path.pop()
```
递归法+回溯（版本二）
```Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
import copy
from typing import List, Optional
class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
if not root:
return []
result = []
self.generate_paths(root, [], result)
return result
def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None:
if not node:
return
path.append(node.val)
if not node.left and not node.right:
result.append('->'.join(map(str, path)))
else:
self.generate_paths(node.left, copy.copy(path), result)
self.generate_paths(node.right, copy.copy(path), result)
path.pop()
```
递归法+隐形回溯
```Python
# Definition for a binary tree node.

62
problems/0344.反转字符串.md
View File

@ -174,6 +174,7 @@ class Solution {
```
Python：
（版本一） 双指针
```python
class Solution:
def reverseString(self, s: List[str]) -> None:
@ -182,15 +183,70 @@ class Solution:
"""
left, right = 0, len(s) - 1
# 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低
# 推荐该写法，更加通俗易懂
# 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(len(s)//2)更低
# 因为while每次循环需要进行条件判断，而range函数不需要，直接生成数字，因此时间复杂度更低。推荐使用range
while left < right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
```
（版本二） 使用栈
```python
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
stack = []
for char in s:
stack.append(char)
for i in range(len(s)):
s[i] = stack.pop()
```
（版本三） 使用range
```python
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
n = len(s)
for i in range(n // 2):
s[i], s[n - i - 1] = s[n - i - 1], s[i]
```
（版本四） 使用reversed
```python
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = reversed(s)
```
（版本五） 使用切片
```python
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[::-1]
```
（版本六） 使用列表推导
```python
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = [s[i] for i in range(len(s) - 1, -1, -1)]
```
Go：
```Go
func reverseString(s []byte) {

34
problems/0349.两个数组的交集.md
View File

@ -160,22 +160,29 @@ class Solution {
```
Python3：
（版本一） 使用字典和集合
```python
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
val_dict = {}
ans = []
# 使用哈希表存储一个数组中的所有元素
table = {}
for num in nums1:
val_dict[num] = 1
for num in nums2:
if num in val_dict.keys() and val_dict[num] == 1:
ans.append(num)
val_dict[num] = 0
table[num] = table.get(num, 0) + 1
return ans
# 使用集合存储结果
res = set()
for num in nums2:
if num in table:
res.add(num)
del table[num]
return list(res)
```
（版本二） 使用数组
```python
class Solution: # 使用数组方法
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
count1 = [0]*1001
count2 = [0]*1001
@ -190,7 +197,14 @@ class Solution: # 使用数组方法
return result
```
（版本三） 使用集合
```python
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
return list(set(nums1) & set(nums2))
```
Go：
```go

99
problems/0383.赎金信.md
View File

@ -146,34 +146,27 @@ class Solution {
```
Python写法一（使用数组作为哈希表）：
（版本一）使用数组
```python
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
arr = [0] * 26
for x in magazine: # 记录 magazine里各个字符出现次数
arr[ord(x) - ord('a')] += 1
for x in ransomNote: # 在arr里对应的字符个数做--操作
if arr[ord(x) - ord('a')] == 0: # 如果没有出现过直接返回
return False
else:
arr[ord(x) - ord('a')] -= 1
return True
ransom_count = [0] * 26
magazine_count = [0] * 26
for c in ransomNote:
ransom_count[ord(c) - ord('a')] += 1
for c in magazine:
magazine_count[ord(c) - ord('a')] += 1
return all(ransom_count[i] <= magazine_count[i] for i in range(26))
```
Python写法二（使用defaultdict）：
（版本二）使用defaultdict
```python
from collections import defaultdict
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
from collections import defaultdict
hashmap = defaultdict(int)
for x in magazine:
@ -181,59 +174,43 @@ class Solution:
for x in ransomNote:
value = hashmap.get(x)
if value is None or value == 0:
if not value or not value:
return False
else:
hashmap[x] -= 1
return True
```
Python写法三：
```python
class Solution(object):
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
# use a dict to store the number of letter occurance in ransomNote
hashmap = dict()
for s in ransomNote:
if s in hashmap:
hashmap[s] += 1
else:
hashmap[s] = 1
# check if the letter we need can be found in magazine
for l in magazine:
if l in hashmap:
hashmap[l] -= 1
for key in hashmap:
if hashmap[key] > 0:
return False
return True
```
Python写法四：
（版本三）使用字典
```python
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
c1 = collections.Counter(ransomNote)
c2 = collections.Counter(magazine)
x = c1 - c2
#x只保留值大于0的符号，当c1里面的符号个数小于c2时，不会被保留
#所以x只保留下了，magazine不能表达的
if(len(x)==0):
return True
else:
return False
counts = {}
for c in magazine:
counts[c] = counts.get(c, 0) + 1
for c in ransomNote:
if c not in counts or counts[c] == 0:
return False
counts[c] -= 1
return True
```
（版本四）使用Counter
```python
from collections import Counter
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
return not Counter(ransomNote) - Counter(magazine)
```
（版本五）使用count
```python
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote))
```
Go：

26
problems/0404.左叶子之和.md
View File

@ -250,19 +250,27 @@ class Solution {
**递归后序遍历**
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sumOfLeftLeaves(self, root: TreeNode) -> int:
if not root:
def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
left_left_leaves_sum = self.sumOfLeftLeaves(root.left) # 左
right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右
cur_left_leaf_val = 0
if root.left and not root.left.left and not root.left.right:
cur_left_leaf_val = root.left.val
# 检查根节点的左子节点是否为叶节点
if root.left and not root.left.left and not root.left.right:
left_val = root.left.val
else:
left_val = self.sumOfLeftLeaves(root.left)
return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中
# 递归地计算右子树左叶节点的和
right_val = self.sumOfLeftLeaves(root.right)
return left_val + right_val
```
**迭代**

36
problems/0454.四数相加II.md
View File

@ -118,11 +118,11 @@ class Solution {
```
Python：
（版本一） 使用字典
```python
class Solution(object):
def fourSumCount(self, nums1, nums2, nums3, nums4):
# use a dict to store the elements in nums1 and nums2 and their sum
# 使用字典存储nums1和nums2中的元素及其和
hashmap = dict()
for n1 in nums1:
for n2 in nums2:
@ -131,7 +131,7 @@ class Solution(object):
else:
hashmap[n1+n2] = 1
# if the -(a+b) exists in nums3 and nums4, we shall add the count
# 如果 -(n1+n2) 存在于nums3和nums4, 存入结果
count = 0
for n3 in nums3:
for n4 in nums4:
@ -142,20 +142,40 @@ class Solution(object):
```
（版本二） 使用字典
```python
class Solution(object):
def fourSumCount(self, nums1, nums2, nums3, nums4):
# 使用字典存储nums1和nums2中的元素及其和
hashmap = dict()
for n1 in nums1:
for n2 in nums2:
hashmap[n1+n2] = hashmap.get(n1+n2, 0) + 1
# 如果 -(n1+n2) 存在于nums3和nums4, 存入结果
count = 0
for n3 in nums3:
for n4 in nums4:
key = - n3 - n4
if key in hashmap:
count += hashmap[key]
return count
```
（版本三）使用 defaultdict
```python
from collections import defaultdict
class Solution:
def fourSumCount(self, nums1: list, nums2: list, nums3: list, nums4: list) -> int:
from collections import defaultdict # You may use normal dict instead.
rec, cnt = defaultdict(lambda : 0), 0
# To store the summary of all the possible combinations of nums1 & nums2, together with their frequencies.
for i in nums1:
for j in nums2:
rec[i+j] += 1
# To add up the frequencies if the corresponding value occurs in the dictionary
for i in nums3:
for j in nums4:
cnt += rec.get(-(i+j), 0) # No matched key, return 0.
cnt += rec.get(-(i+j), 0)
return cnt
```

39
problems/0459.重复的子字符串.md
View File

@ -264,8 +264,7 @@ class Solution {
Python：
这里使用了前缀表统一减一的实现方式
（版本一） 前缀表 减一
```python
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
@ -289,7 +288,7 @@ class Solution:
return nxt
```
前缀表（不减一）的代码实现
（版本二） 前缀表 不减一
```python
class Solution:
@ -314,6 +313,40 @@ class Solution:
return nxt
```
（版本三） 使用 find
```python
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
n = len(s)
if n <= 1:
return False
ss = s[1:] + s[:-1]
print(ss.find(s))
return ss.find(s) != -1
```
（版本四） 暴力法
```python
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
n = len(s)
if n <= 1:
return False
substr = ""
for i in range(1, n//2 + 1):
if n % i == 0:
substr = s[:i]
if substr * (n//i) == s:
return True
return False
```
Go：
这里使用了前缀表统一减一的实现方式

102
problems/0513.找树左下角的值.md
View File

@ -271,52 +271,80 @@ class Solution {
### Python
递归:
（版本一）递归法 + 回溯
```python
class Solution:
def findBottomLeftValue(self, root: TreeNode) -> int:
max_depth = -float("INF")
leftmost_val = 0
self.max_depth = float('-inf')
self.result = None
self.traversal(root, 0)
return self.result
def traversal(self, node, depth):
if not node.left and not node.right:
if depth > self.max_depth:
self.max_depth = depth
self.result = node.val
return
if node.left:
depth += 1
self.traversal(node.left, depth)
depth -= 1
if node.right:
depth += 1
self.traversal(node.right, depth)
depth -= 1
def __traverse(root, cur_depth):
nonlocal max_depth, leftmost_val
if not root.left and not root.right:
if cur_depth > max_depth:
max_depth = cur_depth
leftmost_val = root.val
if root.left:
cur_depth += 1
__traverse(root.left, cur_depth)
cur_depth -= 1
if root.right:
cur_depth += 1
__traverse(root.right, cur_depth)
cur_depth -= 1
__traverse(root, 0)
return leftmost_val
```
迭代 - 层序遍历:
（版本二）递归法+精简
```python
class Solution:
def findBottomLeftValue(self, root: TreeNode) -> int:
queue = deque()
if root:
queue.append(root)
result = 0
while queue:
q_len = len(queue)
for i in range(q_len):
if i == 0:
result = queue[i].val
cur = queue.popleft()
if cur.left:
queue.append(cur.left)
if cur.right:
queue.append(cur.right)
return result
self.max_depth = float('-inf')
self.result = None
self.traversal(root, 0)
return self.result
def traversal(self, node, depth):
if not node.left and not node.right:
if depth > self.max_depth:
self.max_depth = depth
self.result = node.val
return
if node.left:
self.traversal(node.left, depth+1)
if node.right:
self.traversal(node.right, depth+1)
```
(版本三） 迭代法
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
from collections import deque
class Solution:
def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
queue = deque([root])
while queue:
size = len(queue)
leftmost = queue[0].val
for i in range(size):
node = queue.popleft()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if not queue:
return leftmost
```
### Go

21
problems/0647.回文子串.md
View File

@ -267,6 +267,27 @@ class Solution {
return ans;
}
}
```
动态规划：简洁版
```java
class Solution {
public int countSubstrings(String s) {
boolean[][] dp = new boolean[s.length()][s.length()];
int res = 0;
for (int i = s.length() - 1; i >= 0; i--) {
for (int j = i; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j) && (j - i <= 1 || dp[i + 1][j - 1])) {
res++;
dp[i][j] = true;
}
}
}
return res;
}
}
```
中心扩散法：

311
problems/0707.设计链表.md
View File

@ -485,177 +485,176 @@ class MyLinkedList {
Python：
```python
# 单链表
class Node(object):
def __init__(self, x=0):
self.val = x
self.next = None
class MyLinkedList(object):
def __init__(self):
self.head = Node()
self.size = 0 # 设置一个链表长度的属性，便于后续操作，注意每次增和删的时候都要更新
def get(self, index):
"""
:type index: int
:rtype: int
"""
if index < 0 or index >= self.size:
return -1
cur = self.head.next
while(index):
cur = cur.next
index -= 1
return cur.val
def addAtHead(self, val):
"""
:type val: int
:rtype: None
"""
new_node = Node(val)
new_node.next = self.head.next
self.head.next = new_node
self.size += 1
def addAtTail(self, val):
"""
:type val: int
:rtype: None
"""
new_node = Node(val)
cur = self.head
while(cur.next):
cur = cur.next
cur.next = new_node
self.size += 1
def addAtIndex(self, index, val):
"""
:type index: int
:type val: int
:rtype: None
"""
if index < 0:
self.addAtHead(val)
return
elif index == self.size:
self.addAtTail(val)
return
elif index > self.size:
return
node = Node(val)
pre = self.head
while(index):
pre = pre.next
index -= 1
node.next = pre.next
pre.next = node
self.size += 1
def deleteAtIndex(self, index):
"""
:type index: int
:rtype: None
"""
if index < 0 or index >= self.size:
return
pre = self.head
while(index):
pre = pre.next
index -= 1
pre.next = pre.next.next
self.size -= 1
# 双链表
# 相对于单链表, Node新增了prev属性
class Node:
def __init__(self, val):
（版本一）单链表法
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.prev = None
self.next = None
self.next = next
class MyLinkedList:
def __init__(self):
self._head, self._tail = Node(0), Node(0) # 虚拟节点
self._head.next, self._tail.prev = self._tail, self._head
self._count = 0 # 添加的节点数
def _get_node(self, index: int) -> Node:
# 当index小于_count//2时, 使用_head查找更快, 反之_tail更快
if index >= self._count // 2:
# 使用prev往前找
node = self._tail
for _ in range(self._count - index):
node = node.prev
else:
# 使用next往后找
node = self._head
for _ in range(index + 1):
node = node.next
return node
self.dummy_head = ListNode()
self.size = 0
def get(self, index: int) -> int:
"""
Get the value of the index-th node in the linked list. If the index is invalid, return -1.
"""
if 0 <= index < self._count:
node = self._get_node(index)
return node.val
else:
if index < 0 or index >= self.size:
return -1
current = self.dummy_head.next
for i in range(index):
current = current.next
return current.val
def addAtHead(self, val: int) -> None:
"""
Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
"""
self._update(self._head, self._head.next, val)
self.dummy_head.next = ListNode(val, self.dummy_head.next)
self.size += 1
def addAtTail(self, val: int) -> None:
"""
Append a node of value val to the last element of the linked list.
"""
self._update(self._tail.prev, self._tail, val)
current = self.dummy_head
while current.next:
current = current.next
current.next = ListNode(val)
self.size += 1
def addAtIndex(self, index: int, val: int) -> None:
"""
Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
"""
if index < 0:
index = 0
elif index > self._count:
if index < 0 or index > self.size:
return
node = self._get_node(index)
self._update(node.prev, node, val)
def _update(self, prev: Node, next: Node, val: int) -> None:
"""
更新节点
:param prev: 相对于更新的前一个节点
:param next: 相对于更新的后一个节点
:param val: 要添加的节点值
"""
# 计数累加
self._count += 1
node = Node(val)
prev.next, next.prev = node, node
node.prev, node.next = prev, next
current = self.dummy_head
for i in range(index):
current = current.next
current.next = ListNode(val, current.next)
self.size += 1
def deleteAtIndex(self, index: int) -> None:
"""
Delete the index-th node in the linked list, if the index is valid.
"""
if 0 <= index < self._count:
node = self._get_node(index)
# 计数-1
self._count -= 1
node.prev.next, node.next.prev = node.next, node.prev
if index < 0 or index >= self.size:
return
current = self.dummy_head
for i in range(index):
current = current.next
current.next = current.next.next
self.size -= 1
# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)
```
```python
（版本二）双链表法
class ListNode:
def __init__(self, val=0, prev=None, next=None):
self.val = val
self.prev = prev
self.next = next
class MyLinkedList:
def __init__(self):
self.head = None
self.tail = None
self.size = 0
def get(self, index: int) -> int:
if index < 0 or index >= self.size:
return -1
if index < self.size // 2:
current = self.head
for i in range(index):
current = current.next
else:
current = self.tail
for i in range(self.size - index - 1):
current = current.prev
return current.val
def addAtHead(self, val: int) -> None:
new_node = ListNode(val, None, self.head)
if self.head:
self.head.prev = new_node
else:
self.tail = new_node
self.head = new_node
self.size += 1
def addAtTail(self, val: int) -> None:
new_node = ListNode(val, self.tail, None)
if self.tail:
self.tail.next = new_node
else:
self.head = new_node
self.tail = new_node
self.size += 1
def addAtIndex(self, index: int, val: int) -> None:
if index < 0 or index > self.size:
return
if index == 0:
self.addAtHead(val)
elif index == self.size:
self.addAtTail(val)
else:
if index < self.size // 2:
current = self.head
for i in range(index - 1):
current = current.next
else:
current = self.tail
for i in range(self.size - index):
current = current.prev
new_node = ListNode(val, current, current.next)
current.next.prev = new_node
current.next = new_node
self.size += 1
def deleteAtIndex(self, index: int) -> None:
if index < 0 or index >= self.size:
return
if index == 0:
self.head = self.head.next
if self.head:
self.head.prev = None
else:
self.tail = None
elif index == self.size - 1:
self.tail = self.tail.prev
if self.tail:
self.tail.next = None
else:
self.head = None
else:
if index < self.size // 2:
current = self.head
for i in range(index):
current = current.next
else:
current = self.tail
for i in range(self.size - index - 1):
current = current.prev
current.prev.next = current.next
current.next.prev = current.prev
self.size -= 1
# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)
```
Go：

41
problems/0977.有序数组的平方.md
View File

@ -140,22 +140,37 @@ class Solution {
Python：
```Python
（版本一）双指针法
class Solution:
def sortedSquares(self, nums: List[int]) -> List[int]:
n = len(nums)
i,j,k = 0,n - 1,n - 1
ans = [-1] * n
while i <= j:
lm = nums[i] ** 2
rm = nums[j] ** 2
if lm > rm:
ans[k] = lm
i += 1
l, r, i = 0, len(nums)-1, len(nums)-1
res = [float('inf')] * len(nums) # 需要提前定义列表，存放结果
while l <= r:
if nums[l] ** 2 < nums[r] ** 2: # 左右边界进行对比，找出最大值
res[i] = nums[r] ** 2
r -= 1 # 右指针往左移动
else:
ans[k] = rm
j -= 1
k -= 1
return ans
res[i] = nums[l] ** 2
l += 1 # 左指针往右移动
i -= 1 # 存放结果的指针需要往前平移一位
return res
```
```Python
（版本二）暴力排序法
class Solution:
def sortedSquares(self, nums: List[int]) -> List[int]:
for i in range(len(nums)):
nums[i] *= nums[i]
nums.sort()
return nums
```
```Python
（版本三）暴力排序法+列表推导法
class Solution:
def sortedSquares(self, nums: List[int]) -> List[int]:
return sorted(x*x for x in nums)
```
Go：

8
problems/二叉树的迭代遍历.md
View File

@ -258,7 +258,9 @@ class Solution:
if node.left:
stack.append(node.left)
return result
```
```python
# 中序遍历-迭代-LC94_二叉树的中序遍历
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
@ -279,7 +281,9 @@ class Solution:
# 取栈顶元素右结点
cur = cur.right
return result
```
```python
# 后序遍历-迭代-LC145_二叉树的后序遍历
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:

57
problems/二叉树的递归遍历.md
View File

@ -174,50 +174,49 @@ class Solution {
Python：
```python
# 前序遍历-递归-LC144_二叉树的前序遍历
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
# 保存结果
result = []
def traversal(root: TreeNode):
if root == None:
return
result.append(root.val) # 前序
traversal(root.left) # 左
traversal(root.right) # 右
if not root:
return []
traversal(root)
return result
left = self.preorderTraversal(root.left)
right = self.preorderTraversal(root.right)
return [root.val] + left + right
```
```python
# 中序遍历-递归-LC94_二叉树的中序遍历
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
result = []
if root is None:
return []
def traversal(root: TreeNode):
if root == None:
return
traversal(root.left) # 左
result.append(root.val) # 中序
traversal(root.right) # 右
left = self.inorderTraversal(root.left)
right = self.inorderTraversal(root.right)
return left + [root.val] + right
```
```python
traversal(root)
return result
# 后序遍历-递归-LC145_二叉树的后序遍历
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
result = []
if not root:
return []
def traversal(root: TreeNode):
if root == None:
return
traversal(root.left) # 左
traversal(root.right) # 右
result.append(root.val) # 后序
left = self.postorderTraversal(root.left)
right = self.postorderTraversal(root.right)
traversal(root)
return result
return left + right + [root.val]
```
Go：

2
problems/前序/刷力扣用不用库函数.md
View File

@ -24,5 +24,5 @@
例如for循环里套一个字符串的insert，erase之类的操作，你说时间复杂度是多少呢，很明显是O(n^2)的时间复杂度了。
在刷题的时候本着我说的标准来使用库函数，详细对大家回有所帮助！
在刷题的时候本着我说的标准来使用库函数，相信对大家会有所帮助！

33
problems/剑指Offer05.替换空格.md
View File

@ -266,6 +266,8 @@ func replaceSpace(s string) string {
python：
#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能为O(1)
（版本一）转换成列表，并且添加相匹配的空间，然后进行填充
```python
class Solution:
def replaceSpace(self, s: str) -> str:
@ -290,14 +292,22 @@ class Solution:
return ''.join(res)
```
（版本二）添加空列表，添加匹配的结果
```python
class Solution:
def replaceSpace(self, s: str) -> str:
res = []
for i in range(len(s)):
if s[i] == ' ':
res.append('%20')
else:
res.append(s[i])
return ''.join(res)
```
（版本三）使用切片
```python
class Solution:
def replaceSpace(self, s: str) -> str:
# method 1 - Very rude
return "%20".join(s.split(" "))
# method 2 - Reverse the s when counting in for loop, then update from the end.
n = len(s)
for e, i in enumerate(s[::-1]):
print(i, e)
@ -306,7 +316,18 @@ class Solution:
print("")
return s
```
（版本四）使用join + split
```python
class Solution:
def replaceSpace(self, s: str) -> str:
return "%20".join(s.split(" "))
```
（版本五）使用replace
```python
class Solution:
def replaceSpace(self, s: str) -> str:
return s.replace(' ', '%20')
```
javaScript:
```js

54
problems/剑指Offer58-II.左旋转字符串.md
View File

@ -142,15 +142,16 @@ class Solution {
```
python:
（版本一）使用切片
```python
# 方法一：可以使用切片方法
class Solution:
def reverseLeftWords(self, s: str, n: int) -> str:
return s[n:] + s[0:n]
return s[n:] + s[:n]
```
（版本二）使用reversed + join
```python
# 方法二：也可以使用上文描述的方法，有些面试中不允许使用切片，那就使用上文作者提到的方法
class Solution:
def reverseLeftWords(self, s: str, n: int) -> str:
s = list(s)
@ -161,32 +162,29 @@ class Solution:
return "".join(s)
```
（版本三）自定义reversed函数
```python
# 方法三：如果连reversed也不让使用，那么自己手写一个
class Solution:
def reverseLeftWords(self, s: str, n: int) -> str:
def reverse_sub(lst, left, right):
while left < right:
lst[left], lst[right] = lst[right], lst[left]
left += 1
right -= 1
def reverseLeftWords(self, s: str, n: int) -> str:
s_list = list(s)
res = list(s)
end = len(res) - 1
reverse_sub(res, 0, n - 1)
reverse_sub(res, n, end)
reverse_sub(res, 0, end)
return ''.join(res)
self.reverse(s_list, 0, n - 1)
self.reverse(s_list, n, len(s_list) - 1)
self.reverse(s_list, 0, len(s_list) - 1)
# 同方法二
# 时间复杂度：O(n)
# 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改
return ''.join(s_list)
def reverse(self, s, start, end):
while start < end:
s[start], s[end] = s[end], s[start]
start += 1
end -= 1
```
（版本四）使用 模 +下标
```python 3
#方法四：考虑不能用切片的情况下，利用模+下标实现
class Solution:
def reverseLeftWords(self, s: str, n: int) -> str:
new_s = ''
@ -196,17 +194,21 @@ class Solution:
return new_s
```
（版本五）使用 模 + 切片
```python 3
# 方法五：另类的切片方法
class Solution:
def reverseLeftWords(self, s: str, n: int) -> str:
n = len(s)
s = s + s
return s[k : n+k]
l = len(s)
# 复制输入字符串与它自己连接
s = s + s
# 计算旋转字符串的起始索引
k = n % (l * 2)
# 从连接的字符串中提取旋转后的字符串并返回
return s[k : k + l]
# 时间复杂度：O(n)
# 空间复杂度：O(n)
```
Go：

103
problems/面试题02.07.链表相交.md
View File

@ -150,9 +150,13 @@ public class Solution {
}
```
Python：
```python
（版本一）求长度，同时出发
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
lenA, lenB = 0, 0
@ -178,8 +182,105 @@ class Solution:
curB = curB.next
return None
```
```python
（版本二）求长度，同时出发 （代码复用）
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
lenA = self.getLength(headA)
lenB = self.getLength(headB)
# 通过移动较长的链表，使两链表长度相等
if lenA > lenB:
headA = self.moveForward(headA, lenA - lenB)
else:
headB = self.moveForward(headB, lenB - lenA)
# 将两个头向前移动，直到它们相交
while headA and headB:
if headA == headB:
return headA
headA = headA.next
headB = headB.next
return None
def getLength(self, head: ListNode) -> int:
length = 0
while head:
length += 1
head = head.next
return length
def moveForward(self, head: ListNode, steps: int) -> ListNode:
while steps > 0:
head = head.next
steps -= 1
return head
```
```python
（版本三）求长度，同时出发 （代码复用 + 精简）
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
dis = self.getLength(headA) - self.getLength(headB)
# 通过移动较长的链表，使两链表长度相等
if dis > 0:
headA = self.moveForward(headA, dis)
else:
headB = self.moveForward(headB, abs(dis))
# 将两个头向前移动，直到它们相交
while headA and headB:
if headA == headB:
return headA
headA = headA.next
headB = headB.next
return None
def getLength(self, head: ListNode) -> int:
length = 0
while head:
length += 1
head = head.next
return length
def moveForward(self, head: ListNode, steps: int) -> ListNode:
while steps > 0:
head = head.next
steps -= 1
return head
```
```python
（版本四）等比例法
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
Go：
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
# 处理边缘情况
if not headA or not headB:
return None
# 在每个链表的头部初始化两个指针
pointerA = headA
pointerB = headB
# 遍历两个链表直到指针相交
while pointerA != pointerB:
# 将指针向前移动一个节点
pointerA = pointerA.next if pointerA else headB
pointerB = pointerB.next if pointerB else headA
# 如果相交，指针将位于交点节点，如果没有交点，值为None
return pointerA
```
Go:
```go
func getIntersectionNode(headA, headB *ListNode) *ListNode {