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Merge pull request #2078 from jianghongcheng/master

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python 二叉树 章节 持续更新。
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程序员Carl
2023-05-07 09:58:39 +08:00
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commit c80796ce26
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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：

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：

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：

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 {

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

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：

36
problems/面试题02.07.链表相交.md
View File

@ -214,7 +214,41 @@ class Solution:
return head
```
```python
（版本三）等比例法
（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):