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Merge branch 'master' into master

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This commit is contained in:
Carl Sun
2021-06-02 20:38:29 +08:00
committed by GitHub
parent 3362183d84 cd18cb90f1
commit e35169f11a
7 changed files with 332 additions and 53 deletions
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48
problems/0101.对称二叉树.md
View File

@ -363,6 +363,54 @@ Python：
Go：
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
// 递归
func defs(left *TreeNode, right *TreeNode) bool {
if left == nil && right == nil {
return true;
};
if left == nil || right == nil {
return false;
};
if left.Val != right.Val {
return false;
}
return defs(left.Left, right.Right) && defs(right.Left, left.Right);
}
func isSymmetric(root *TreeNode) bool {
return defs(root.Left, root.Right);
}
// 迭代
func isSymmetric(root *TreeNode) bool {
var queue []*TreeNode;
if root != nil {
queue = append(queue, root.Left, root.Right);
}
for len(queue) > 0 {
left := queue[0];
right := queue[1];
queue = queue[2:];
if left == nil && right == nil {
continue;
}
if left == nil || right == nil || left.Val != right.Val {
return false;
};
queue = append(queue, left.Left, right.Right, right.Left, left.Right);
}
return true;
}
```
JavaScript
```javascript

49
problems/0104.二叉树的最大深度.md
View File

@ -284,6 +284,55 @@ Python：
Go：
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func max (a, b int) int {
if a > b {
return a;
}
return b;
}
// 递归
func maxDepth(root *TreeNode) int {
if root == nil {
return 0;
}
return max(maxDepth(root.Left), maxDepth(root.Right)) + 1;
}
// 遍历
func maxDepth(root *TreeNode) int {
levl := 0;
queue := make([]*TreeNode, 0);
if root != nil {
queue = append(queue, root);
}
for l := len(queue); l > 0; {
for ;l > 0;l-- {
node := queue[0];
if node.Left != nil {
queue = append(queue, node.Left);
}
if node.Right != nil {
queue = append(queue, node.Right);
}
queue = queue[1:];
}
levl++;
l = len(queue);
}
return levl;
}
```
JavaScript
```javascript

58
problems/0111.二叉树的最小深度.md
View File

@ -301,6 +301,64 @@ class Solution:
Go：
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func min(a, b int) int {
if a < b {
return a;
}
return b;
}
// 递归
func minDepth(root *TreeNode) int {
if root == nil {
return 0;
}
if root.Left == nil && root.Right != nil {
return 1 + minDepth(root.Right);
}
if root.Right == nil && root.Left != nil {
return 1 + minDepth(root.Left);
}
return min(minDepth(root.Left), minDepth(root.Right)) + 1;
}
// 迭代
func minDepth(root *TreeNode) int {
dep := 0;
queue := make([]*TreeNode, 0);
if root != nil {
queue = append(queue, root);
}
for l := len(queue); l > 0; {
dep++;
for ; l > 0; l-- {
node := queue[0];
if node.Left == nil && node.Right == nil {
return dep;
}
if node.Left != nil {
queue = append(queue, node.Left);
}
if node.Right != nil {
queue = append(queue, node.Right);
}
queue = queue[1:];
}
l = len(queue);
}
return dep;
}
```
JavaScript:

35
problems/0239.滑动窗口最大值.md
View File

@ -263,6 +263,41 @@ class Solution {
```
Python：
```python
class MyQueue: #单调队列（从大到小
def __init__(self):
self.queue = [] #使用list来实现单调队列
#每次弹出的时候，比较当前要弹出的数值是否等于队列出口元素的数值，如果相等则弹出。
#同时pop之前判断队列当前是否为空。
def pop(self, value):
if self.queue and value == self.queue[0]:
self.queue.pop(0)#list.pop()时间复杂度为O(n),这里可以使用collections.deque()
#如果push的数值大于入口元素的数值，那么就将队列后端的数值弹出，直到push的数值小于等于队列入口元素的数值为止。
#这样就保持了队列里的数值是单调从大到小的了。
def push(self, value):
while self.queue and value > self.queue[-1]:
self.queue.pop()
self.queue.append(value)
#查询当前队列里的最大值 直接返回队列前端也就是front就可以了。
def front(self):
return self.queue[0]
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
que = MyQueue()
result = []
for i in range(k): #先将前k的元素放进队列
que.push(nums[i])
result.append(que.front()) #result 记录前k的元素的最大值
for i in range(k, len(nums)):
que.pop(nums[i - k]) #滑动窗口移除最前面元素
que.push(nums[i]) #滑动窗口前加入最后面的元素
result.append(que.front()) #记录对应的最大值
return result
```
Go：

75
problems/0347.前K个高频元素.md
View File

@ -163,64 +163,33 @@ class Solution {
Python：
```python
#时间复杂度：O(nlogk)
#空间复杂度：O(n)
import heapq
class Solution:
def sift(self, alist, low, high):
'''小根堆构建'''
i = low
j = 2 * i + 1
tmp = alist[low]
while j <= high:
if j + 1 <= high and alist[j+1] <= alist[j]:
j += 1
if alist[j] < tmp:
alist[i] = alist[j]
i = j
j = 2 * i + 1
else:
alist[i] = tmp
break
else:
alist[i] = tmp
def topK(self, nums, k):
# 建立小根堆
heap = nums[:k]
for i in range((k-2)//2, -1, -1):
self.sift(heap, i, k-1)
# 把后续的k到len(nums)填充到小根堆里
for i in range(k, len(nums)):
if nums[i] >= heap[0]:
heap[0] = nums[i]
self.sift(heap, 0, k-1)
# 排序
for i in range(k-1, -1, -1):
heap[0], heap[i]= heap[i], heap[0]
self.sift(heap, 0, i-1)
return heap
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
dict1 = dict()
for val in nums:
if val not in dict1:
dict1[val] = 1
else:
dict1[val] += 1
res = list()
ind = list()
for item in dict1:
res.append([dict1[item], item])
result = list()
heap = self.topK(res, k)
print(heap)
for val in heap:
result.append(val[1])
#要统计元素出现频率
map_ = {} #nums[i]:对应出现的次数
for i in range(len(nums)):
map_[nums[i]] = map_.get(nums[i], 0) + 1
#对频率排序
#定义一个小顶堆，大小为k
pri_que = [] #小顶堆
#用固定大小为k的小顶堆，扫面所有频率的数值
for key, freq in map_.items():
heapq.heappush(pri_que, (freq, key))
if len(pri_que) > k: #如果堆的大小大于了K，则队列弹出，保证堆的大小一直为k
heapq.heappop(pri_que)
#找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒叙来输出到数组
result = [0] * k
for i in range(k-1, -1, -1):
result[i] = heapq.heappop(pri_que)[1]
return result
```
Go：

49
problems/0459.重复的子字符串.md
View File

@ -184,6 +184,55 @@ class Solution {
Python：
这里使用了前缀表统一减一的实现方式
```python
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
if len(s) == 0:
return False
nxt = [0] * len(s)
self.getNext(nxt, s)
if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0:
return True
return False
def getNext(self, nxt, s):
nxt[0] = -1
j = -1
for i in range(1, len(s)):
while j >= 0 and s[i] != s[j+1]:
j = nxt[j]
if s[i] == s[j+1]:
j += 1
nxt[i] = j
return nxt
```
前缀表（不减一）的代码实现
```python
class Solution:
def repeatedSubstringPattern(self, s: str) -> bool:
if len(s) == 0:
return False
nxt = [0] * len(s)
self.getNext(nxt, s)
if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0:
return True
return False
def getNext(self, nxt, s):
nxt[0] = 0
j = 0
for i in range(1, len(s)):
while j > 0 and s[i] != s[j]:
j = nxt[j - 1]
if s[i] == s[j]:
j += 1
nxt[i] = j
return nxt
```
Go：

71
problems/二叉树的统一迭代法.md
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@ -239,7 +239,78 @@ Java：
```
Python：
> 迭代法前序遍历
```python
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
result = []
st= []
if root:
st.append(root)
while st:
node = st.pop()
if node != None:
if node.right: #右
st.append(node.right)
if node.left: #左
st.append(node.left)
st.append(node) #中
st.append(None)
else:
node = st.pop()
result.append(node.val)
return result
```
> 迭代法中序遍历
```python
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
result = []
st = []
if root:
st.append(root)
while st:
node = st.pop()
if node != None:
if node.right: #添加右节点（空节点不入栈）
st.append(node.right)
st.append(node) #添加中节点
st.append(None) #中节点访问过，但是还没有处理，加入空节点做为标记。
if node.left: #添加左节点（空节点不入栈）
st.append(node.left)
else: #只有遇到空节点的时候，才将下一个节点放进结果集
node = st.pop() #重新取出栈中元素
result.append(node.val) #加入到结果集
return result
```
> 迭代法后序遍历
```python
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
result = []
st = []
if root:
st.append(root)
while st:
node = st.pop()
if node != None:
st.append(node) #中
st.append(None)
if node.right: #右
st.append(node.right)
if node.left: #左
st.append(node.left)
else:
node = st.pop()
result.append(node.val)
return result
```
Go：
> 前序遍历统一迭代法