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Merge branch 'master' of github.com:youngyangyang04/leetcode-master

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youngyangyang04
2021-09-23 10:58:56 +08:00
parent 3ed64ced80 5ed55ffb26
commit 9fa333e646
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2
problems/0027.移除元素.md
View File

@ -185,7 +185,7 @@ func removeElement(nums []int, val int) int {
```
JavaScript:
```
```javascript
//时间复杂度O(n)
//空间复杂度O(1)
var removeElement = (nums, val) => {

28
problems/0077.组合.md
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@ -368,6 +368,34 @@ class Solution {
}
```
Python2:
```python
class Solution(object):
def combine(self, n, k):
"""
:type n: int
:type k: int
:rtype: List[List[int]]
"""
result = []
path = []
def backtracking(n, k, startidx):
if len(path) == k:
result.append(path[:])
return
# 剪枝， 最后k - len(path)个节点直接构造结果，无需递归
last_startidx = n - (k - len(path)) + 1
result.append(path + [idx for idx in range(last_startidx, n + 1)])
for x in range(startidx, last_startidx):
path.append(x)
backtracking(n, k, x + 1) # 递归
path.pop() # 回溯
backtracking(n, k, 1)
return result
```
## Python
```python

57
problems/0078.子集.md
View File

@ -263,6 +263,63 @@ var subsets = function(nums) {
};
```
C:
```c
int* path;
int pathTop;
int** ans;
int ansTop;
//记录二维数组中每个一维数组的长度
int* length;
//将当前path数组复制到ans中
void copy() {
int* tempPath = (int*)malloc(sizeof(int) * pathTop);
int i;
for(i = 0; i < pathTop; i++) {
tempPath[i] = path[i];
}
ans = (int**)realloc(ans, sizeof(int*) * (ansTop+1));
length[ansTop] = pathTop;
ans[ansTop++] = tempPath;
}
void backTracking(int* nums, int numsSize, int startIndex) {
//收集子集，要放在终止添加的上面，否则会漏掉自己
copy();
//若startIndex大于数组大小，返回
if(startIndex >= numsSize) {
return;
}
int j;
for(j = startIndex; j < numsSize; j++) {
//将当前下标数字放入path中
path[pathTop++] = nums[j];
backTracking(nums, numsSize, j+1);
//回溯
pathTop--;
}
}
int** subsets(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
//初始化辅助变量
path = (int*)malloc(sizeof(int) * numsSize);
ans = (int**)malloc(0);
length = (int*)malloc(sizeof(int) * 1500);
ansTop = pathTop = 0;
//进入回溯
backTracking(nums, numsSize, 0);
//设置二维数组中元素个数
*returnSize = ansTop;
//设置二维数组中每个一维数组的长度
*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
int i;
for(i = 0; i < ansTop; i++) {
(*returnColumnSizes)[i] = length[i];
}
return ans;
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

76
problems/0093.复原IP地址.md
View File

@ -476,6 +476,82 @@ func isNormalIp(s string,startIndex,end int)bool{
```
C:
```c
//记录结果
char** result;
int resultTop;
//记录应该加入'.'的位置
int segments[3];
int isValid(char* s, int start, int end) {
if(start > end)
return 0;
if (s[start] == '0' && start != end) { // 0开头的数字不合法
return false;
}
int num = 0;
for (int i = start; i <= end; i++) {
if (s[i] > '9' || s[i] < '0') { // 遇到非数字字符不合法
return false;
}
num = num * 10 + (s[i] - '0');
if (num > 255) { // 如果大于255了不合法
return false;
}
}
return true;
}
//startIndex为起始搜索位置，pointNum为'.'对象
void backTracking(char* s, int startIndex, int pointNum) {
//若'.'数量为3，分隔结束
if(pointNum == 3) {
//若最后一段字符串符合要求，将当前的字符串放入result种
if(isValid(s, startIndex, strlen(s) - 1)) {
char* tempString = (char*)malloc(sizeof(char) * strlen(s) + 4);
int j;
//记录添加字符时tempString的下标
int count = 0;
//记录添加字符时'.'的使用数量
int count1 = 0;
for(j = 0; j < strlen(s); j++) {
tempString[count++] = s[j];
//若'.'的使用数量小于3且当前下标等于'.'下标，添加'.'到数组
if(count1 < 3 && j == segments[count1]) {
tempString[count++] = '.';
count1++;
}
}
tempString[count] = 0;
//扩容result数组
result = (char**)realloc(result, sizeof(char*) * (resultTop + 1));
result[resultTop++] = tempString;
}
return ;
}
int i;
for(i = startIndex; i < strlen(s); i++) {
if(isValid(s, startIndex, i)) {
//记录应该添加'.'的位置
segments[pointNum] = i;
backTracking(s, i + 1, pointNum + 1);
}
else {
break;
}
}
}
char ** restoreIpAddresses(char * s, int* returnSize){
result = (char**)malloc(0);
resultTop = 0;
backTracking(s, 0, 0);
*returnSize = resultTop;
return result;
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

3
problems/0101.对称二叉树.md
View File

@ -185,7 +185,8 @@ public:
queue<TreeNode*> que;
que.push(root->left); // 将左子树头结点加入队列
que.push(root->right); // 将右子树头结点加入队列
while (!que.empty()) { // 接下来就要判断这这两个树是否相互翻转
while (!que.empty()) { // 接下来就要判断这两个树是否相互翻转
TreeNode* leftNode = que.front(); que.pop();
TreeNode* rightNode = que.front(); que.pop();
if (!leftNode && !rightNode) { // 左节点为空、右节点为空，此时说明是对称的

25
problems/0188.买卖股票的最佳时机IV.md
View File

@ -280,6 +280,7 @@ Go：
Javascript：
```javascript
// 方法一：动态规划
const maxProfit = (k,prices) => {
if (prices == null || prices.length < 2 || k == 0) {
return 0;
@ -300,6 +301,30 @@ const maxProfit = (k,prices) => {
return dp[prices.length - 1][2 * k];
};
// 方法二：动态规划+空间优化
var maxProfit = function(k, prices) {
let n = prices.length;
let dp = new Array(2*k+1).fill(0);
// dp 买入状态初始化
for (let i = 1; i <= 2*k; i += 2) {
dp[i] = - prices[0];
}
for (let i = 1; i < n; i++) {
for (let j = 1; j < 2*k+1; j++) {
// j 为奇数：买入状态
if (j % 2) {
dp[j] = Math.max(dp[j], dp[j-1] - prices[i]);
} else {
// j为偶数：卖出状态
dp[j] = Math.max(dp[j], dp[j-1] + prices[i]);
}
}
}
return dp[2*k];
};
```
-----------------------

38
problems/0202.快乐数.md
View File

@ -159,26 +159,28 @@ func getSum(n int) int {
javaScript:
```js
function getN(n) {
if (n == 1 || n == 0) return n;
let res = 0;
while (n) {
res += (n % 10) * (n % 10);
n = parseInt(n / 10);
}
return res;
}
var isHappy = function (n) {
const sumSet = new Set();
while (n != 1 && !sumSet.has(n)) {
sumSet.add(n);
n = getN(n);
}
return n == 1;
};
let m = new Map()
// 使用环形链表的思想 说明出现闭环 退出循环
const getSum = (num) => {
let sum = 0
while (n) {
sum += (n % 10) ** 2
n = Math.floor(n / 10)
}
return sum
}
while (true) {
// n出现过，证明已陷入无限循环
if (m.has(n)) return false
if (n === 1) return true
m.set(n, 1)
n = getSum(n)
}
}
// 方法二：使用环形链表的思想 说明出现闭环 退出循环
var isHappy = function(n) {
if (getN(n) == 1) return true;
let a = getN(n), b = getN(getN(n));

23
problems/0235.二叉搜索树的最近公共祖先.md
View File

@ -268,17 +268,30 @@ class Solution {
递归法：
```python
class Solution:
"""二叉搜索树的最近公共祖先 递归法"""
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root: return root //中
if root.val > p.val and root.val > q.val:
return self.lowestCommonAncestor(root.left,p,q) //左
elif root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right,p,q) //右
else: return root
return self.lowestCommonAncestor(root.left, p, q)
if root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right, p, q)
return root
```
迭代法：
```python
class Solution:
"""二叉搜索树的最近公共祖先 迭代法"""
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
while True:
if root.val > p.val and root.val > q.val:
root = root.left
elif root.val < p.val and root.val < q.val:
root = root.right
else:
return root
```
## Go

21
problems/0236.二叉树的最近公共祖先.md
View File

@ -264,16 +264,21 @@ class Solution {
## Python
```python
//递归
class Solution:
"""二叉树的最近公共祖先 递归法"""
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root or root == p or root == q: return root //找到了节点p或者q，或者遇到空节点
left = self.lowestCommonAncestor(root.left,p,q) //左
right = self.lowestCommonAncestor(root.right,p,q) //右
if left and right: return root //中: left和right不为空，root就是最近公共节点
elif left and not right: return left //目标节点是通过left返回的
elif not left and right: return right //目标节点是通过right返回的
else: return None //没找到
if not root or root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
if left:
return left
return right
```
## Go

23
problems/0455.分发饼干.md
View File

@ -217,6 +217,29 @@ var findContentChildren = function(g, s) {
```
C:
```c
int cmp(int* a, int* b) {
return *a - *b;
}
int findContentChildren(int* g, int gSize, int* s, int sSize){
if(sSize == 0)
return 0;
//将两个数组排序为升序
qsort(g, gSize, sizeof(int), cmp);
qsort(s, sSize, sizeof(int), cmp);
int numFedChildren = 0;
int i = 0;
for(i = 0; i < sSize; ++i) {
if(numFedChildren < gSize && g[numFedChildren] <= s[i])
numFedChildren++;
}
return numFedChildren;
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

21
problems/0718.最长重复子数组.md
View File

@ -278,7 +278,26 @@ const findLength = (A, B) => {
return res;
};
```
> 滚动数组
```javascript
const findLength = (nums1, nums2) => {
let len1 = nums1.length, len2 = nums2.length;
// dp[i][j]: 以nums1[i-1]、nums2[j-1]为结尾的最长公共子数组的长度
let dp = new Array(len2+1).fill(0);
let res = 0;
for (let i = 1; i <= len1; i++) {
for (let j = len2; j > 0; j--) {
if (nums1[i-1] === nums2[j-1]) {
dp[j] = dp[j-1] + 1;
} else {
dp[j] = 0;
}
res = Math.max(res, dp[j]);
}
}
return res;
}
```
-----------------------

30
problems/0739.每日温度.md
View File

@ -276,6 +276,36 @@ func dailyTemperatures(num []int) []int {
}
```
JavaScript：
```javascript
/**
* @param {number[]} temperatures
* @return {number[]}
*/
var dailyTemperatures = function(temperatures) {
let n = temperatures.length;
let res = new Array(n).fill(0);
let stack = []; // 递减栈：用于存储元素右面第一个比他大的元素下标
stack.push(0);
for (let i = 1; i < n; i++) {
// 栈顶元素
let top = stack[stack.length - 1];
if (temperatures[i] < temperatures[top]) {
stack.push(i);
} else if (temperatures[i] === temperatures[top]) {
stack.push(i);
} else {
while (stack.length && temperatures[i] > temperatures[stack[stack.length - 1]]) {
let top = stack.pop();
res[top] = i - top;
}
stack.push(i);
}
}
return res;
};
```
-----------------------

99
problems/0841.钥匙和房间.md
View File

@ -121,10 +121,109 @@ public:
Java：
```java
class Solution {
private void dfs(int key, List<List<Integer>> rooms, List<Boolean> visited) {
if (visited.get(key)) {
return;
}
visited.set(key, true);
for (int k : rooms.get(key)) {
// 深度优先搜索遍历
dfs(k, rooms, visited);
}
}
public boolean canVisitAllRooms(List<List<Integer>> rooms) {
List<Boolean> visited = new ArrayList<Boolean>(){{
for(int i = 0 ; i < rooms.size(); i++){
add(false);
}
}};
dfs(0, rooms, visited);
//检查是否都访问到了
for (boolean flag : visited) {
if (!flag) {
return false;
}
}
return true;
}
}
```
Python：
python3
```python
class Solution:
def dfs(self, key: int, rooms: List[List[int]] , visited : List[bool] ) :
if visited[key] :
return
visited[key] = True
keys = rooms[key]
for i in range(len(keys)) :
# 深度优先搜索遍历
self.dfs(keys[i], rooms, visited)
def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
visited = [False for i in range(len(rooms))]
self.dfs(0, rooms, visited)
# 检查是否都访问到了
for i in range(len(visited)):
if not visited[i] :
return False
return True
```
Go：
```go
func dfs(key int, rooms [][]int, visited []bool ) {
if visited[key] {
return;
}
visited[key] = true
keys := rooms[key]
for _ , key := range keys {
// 深度优先搜索遍历
dfs(key, rooms, visited);
}
}
func canVisitAllRooms(rooms [][]int) bool {
visited := make([]bool, len(rooms));
dfs(0, rooms, visited);
//检查是否都访问到了
for i := 0; i < len(visited); i++ {
if !visited[i] {
return false;
}
}
return true;
}
```
JavaScript：
-----------------------

41
problems/0844.比较含退格的字符串.md
View File

@ -188,8 +188,49 @@ class Solution {
Python：
python3
```python
class Solution:
def get_string(self, s: str) -> str :
bz = []
for i in range(len(s)) :
c = s[i]
if c != '#' :
bz.append(c) # 模拟入栈
elif len(bz) > 0: # 栈非空才能弹栈
bz.pop() # 模拟弹栈
return str(bz)
def backspaceCompare(self, s: str, t: str) -> bool:
return self.get_string(s) == self.get_string(t)
pass
```
Go：
```go
func getString(s string) string {
bz := []rune{}
for _, c := range s {
if c != '#' {
bz = append(bz, c); // 模拟入栈
} else if len(bz) > 0 { // 栈非空才能弹栈
bz = bz[:len(bz)-1] // 模拟弹栈
}
}
return string(bz)
}
func backspaceCompare(s string, t string) bool {
return getString(s) == getString(t)
}
```
JavaScript：
-----------------------

28
problems/0922.按奇偶排序数组II.md
View File

@ -11,6 +11,8 @@
# 922. 按奇偶排序数组II
[力扣题目链接](https://leetcode-cn.com/problems/sort-array-by-parity-ii/)
给定一个非负整数数组 A， A 中一半整数是奇数，一半整数是偶数。
对数组进行排序，以便当 A[i] 为奇数时，i 也是奇数；当 A[i] 为偶数时， i 也是偶数。
@ -147,9 +149,9 @@ class Solution {
}
```
## Python
## Python3
```python3
```python
#方法2
class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
@ -180,6 +182,28 @@ class Solution:
## Go
```go
// 方法一
func sortArrayByParityII(nums []int) []int {
// 分别存放 nums 中的奇数、偶数
even, odd := []int{}, []int{}
for i := 0; i < len(nums); i++ {
if (nums[i] % 2 == 0) {
even = append(even, nums[i])
} else {
odd = append(odd, nums[i])
}
}
// 把奇偶数组重新存回 nums
result := make([]int, len(nums))
index := 0
for i := 0; i < len(even); i++ {
result[index] = even[i]; index++;
result[index] = odd[i]; index++;
}
return result;
}
```
## JavaScript

24
problems/0925.长按键入.md
View File

@ -155,8 +155,32 @@ class Solution:
else: return False
return True
```
Go：
```go
func isLongPressedName(name string, typed string) bool {
if(name[0] != typed[0] || len(name) > len(typed)) {
return false;
}
idx := 0 // name的索引
var last byte // 上个匹配字符
for i := 0; i < len(typed); i++ {
if idx < len(name) && name[idx] == typed[i] {
last = name[idx]
idx++
} else if last == typed[i] {
continue
} else {
return false
}
}
return idx == len(name)
}
```
JavaScript：
-----------------------

40
problems/0941.有效的山脉数组.md
View File

@ -103,14 +103,52 @@ class Solution {
}
```
## Python
## Python3
```python
class Solution:
def validMountainArray(self, arr: List[int]) -> bool:
if len(arr) < 3 :
return False
i = 1
flagIncrease = False # 上升
flagDecrease = False # 下降
while i < len(arr) and arr[i-1] < arr[i]:
flagIncrease = True
i += 1
while i < len(arr) and arr[i-1] > arr[i]:
flagDecrease = True
i += 1
return i == len(arr) and flagIncrease and flagDecrease
```
## Go
```go
func validMountainArray(arr []int) bool {
if len(arr) < 3 {
return false
}
i := 1
flagIncrease := false // 上升
flagDecrease := false // 下降
for ; i < len(arr) && arr[i-1] < arr[i]; i++ {
flagIncrease = true;
}
for ; i < len(arr) && arr[i-1] > arr[i]; i++ {
flagDecrease = true;
}
return i == len(arr) && flagIncrease && flagDecrease;
}
```
## JavaScript

2
problems/数组理论基础.md
View File

@ -88,7 +88,7 @@ int main() {
**所以可以看出在C++中二维数组在地址空间上是连续的**。
像Java是没有指针的，同时也不对程序员暴漏其元素的地址，寻址操作完全交给虚拟机。
像Java是没有指针的，同时也不对程序员暴露其元素的地址，寻址操作完全交给虚拟机。
所以看不到每个元素的地址情况，这里我以Java为例，也做一个实验。

3
problems/背包理论基础01背包-1.md
View File

@ -315,7 +315,6 @@ Python：
def test_2_wei_bag_problem1(bag_size, weight, value) -> int:
rows, cols = len(weight), bag_size + 1
dp = [[0 for _ in range(cols)] for _ in range(rows)]
res = 0
# 初始化dp数组.
for i in range(rows):
@ -334,8 +333,6 @@ def test_2_wei_bag_problem1(bag_size, weight, value) -> int:
else:
# 定义dp数组: dp[i][j] 前i个物品里，放进容量为j的背包，价值总和最大是多少。
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - cur_weight]+ cur_val)
if dp[i][j] > res:
res = dp[i][j]
print(dp)