mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-09 03:34:02 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
Yukun J
@ -263,21 +263,38 @@ public:
|
||||
Java：
|
||||
|
||||
```java
|
||||
public int[] twoSum(int[] nums, int target) {
|
||||
int[] res = new int[2];
|
||||
if(nums == null || nums.length == 0){
|
||||
return res;
|
||||
}
|
||||
Map<Integer, Integer> map = new HashMap<>();
|
||||
for(int i = 0; i < nums.length; i++){
|
||||
int temp = target - nums[i];
|
||||
if(map.containsKey(temp)){
|
||||
res[1] = i;
|
||||
res[0] = map.get(temp);
|
||||
// 双指针 动态规划
|
||||
class Solution {
|
||||
public String longestPalindrome(String s) {
|
||||
if (s.length() == 0 || s.length() == 1) return s;
|
||||
int length = 1;
|
||||
int index = 0;
|
||||
boolean[][] palindrome = new boolean[s.length()][s.length()];
|
||||
for (int i = 0; i < s.length(); i++) {
|
||||
palindrome[i][i] = true;
|
||||
}
|
||||
map.put(nums[i], i);
|
||||
|
||||
for (int L = 2; L <= s.length(); L++) {
|
||||
for (int i = 0; i < s.length(); i++) {
|
||||
int j = i + L - 1;
|
||||
if (j >= s.length()) break;
|
||||
if (s.charAt(i) != s.charAt(j)) {
|
||||
palindrome[i][j] = false;
|
||||
} else {
|
||||
if (j - i < 3) {
|
||||
palindrome[i][j] = true;
|
||||
} else {
|
||||
palindrome[i][j] = palindrome[i + 1][j - 1];
|
||||
}
|
||||
}
|
||||
if (palindrome[i][j] && j - i + 1 > length) {
|
||||
length = j - i + 1;
|
||||
index = i;
|
||||
}
|
||||
}
|
||||
}
|
||||
return s.substring(index, index + length);
|
||||
}
|
||||
return res;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
@ -222,56 +222,6 @@ public:
|
||||
|
||||
## 其他语言补充
|
||||
|
||||
|
||||
### Python
|
||||
|
||||
```python
|
||||
class Solution:
|
||||
def solveNQueens(self, n: int) -> List[List[str]]:
|
||||
if not n: return []
|
||||
board = [['.'] * n for _ in range(n)]
|
||||
res = []
|
||||
def isVaild(board,row, col):
|
||||
#判断同一列是否冲突
|
||||
for i in range(len(board)):
|
||||
if board[i][col] == 'Q':
|
||||
return False
|
||||
# 判断左上角是否冲突
|
||||
i = row -1
|
||||
j = col -1
|
||||
while i>=0 and j>=0:
|
||||
if board[i][j] == 'Q':
|
||||
return False
|
||||
i -= 1
|
||||
j -= 1
|
||||
# 判断右上角是否冲突
|
||||
i = row - 1
|
||||
j = col + 1
|
||||
while i>=0 and j < len(board):
|
||||
if board[i][j] == 'Q':
|
||||
return False
|
||||
i -= 1
|
||||
j += 1
|
||||
return True
|
||||
|
||||
def backtracking(board, row, n):
|
||||
# 如果走到最后一行，说明已经找到一个解
|
||||
if row == n:
|
||||
temp_res = []
|
||||
for temp in board:
|
||||
temp_str = "".join(temp)
|
||||
temp_res.append(temp_str)
|
||||
res.append(temp_res)
|
||||
for col in range(n):
|
||||
if not isVaild(board, row, col):
|
||||
continue
|
||||
board[row][col] = 'Q'
|
||||
backtracking(board, row+1, n)
|
||||
board[row][col] = '.'
|
||||
backtracking(board, 0, n)
|
||||
return res
|
||||
```
|
||||
|
||||
### Java
|
||||
|
||||
```java
|
||||
@ -341,6 +291,55 @@ class Solution {
|
||||
}
|
||||
```
|
||||
|
||||
### Python
|
||||
|
||||
```python
|
||||
class Solution:
|
||||
def solveNQueens(self, n: int) -> List[List[str]]:
|
||||
if not n: return []
|
||||
board = [['.'] * n for _ in range(n)]
|
||||
res = []
|
||||
def isVaild(board,row, col):
|
||||
#判断同一列是否冲突
|
||||
for i in range(len(board)):
|
||||
if board[i][col] == 'Q':
|
||||
return False
|
||||
# 判断左上角是否冲突
|
||||
i = row -1
|
||||
j = col -1
|
||||
while i>=0 and j>=0:
|
||||
if board[i][j] == 'Q':
|
||||
return False
|
||||
i -= 1
|
||||
j -= 1
|
||||
# 判断右上角是否冲突
|
||||
i = row - 1
|
||||
j = col + 1
|
||||
while i>=0 and j < len(board):
|
||||
if board[i][j] == 'Q':
|
||||
return False
|
||||
i -= 1
|
||||
j += 1
|
||||
return True
|
||||
|
||||
def backtracking(board, row, n):
|
||||
# 如果走到最后一行，说明已经找到一个解
|
||||
if row == n:
|
||||
temp_res = []
|
||||
for temp in board:
|
||||
temp_str = "".join(temp)
|
||||
temp_res.append(temp_str)
|
||||
res.append(temp_res)
|
||||
for col in range(n):
|
||||
if not isVaild(board, row, col):
|
||||
continue
|
||||
board[row][col] = 'Q'
|
||||
backtracking(board, row+1, n)
|
||||
board[row][col] = '.'
|
||||
backtracking(board, 0, n)
|
||||
return res
|
||||
```
|
||||
|
||||
|
||||
### Go
|
||||
```Go
|
||||
@ -396,6 +395,8 @@ func isValid(n, row, col int, chessboard [][]string) bool {
|
||||
return true
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
### Javascript
|
||||
```Javascript
|
||||
var solveNQueens = function(n) {
|
||||
|
||||
@ -187,12 +187,14 @@ func climbStairs(n int) int {
|
||||
JavaScript:
|
||||
```javascript
|
||||
var climbStairs = function(n) {
|
||||
const dp = new Array(n+1).fill(0);
|
||||
const weight = [1,2];
|
||||
const dp = new Array(n + 1).fill(0);
|
||||
const m = 2;
|
||||
dp[0] = 1;
|
||||
for(let i = 0; i <= n; i++){ //先遍历背包
|
||||
for(let j = 0; j < weight.length; j++){ // 再遍历物品
|
||||
if(i >= weight[j]) dp[i] += dp[i-weight[j]];
|
||||
for(let i = 1; i <= n; i++){
|
||||
for(let j = 1; j <= m; j++){
|
||||
if(i >= j) {
|
||||
dp[i] += dp[i - j];
|
||||
}
|
||||
}
|
||||
}
|
||||
return dp[n];
|
||||
|
||||
@ -20,7 +20,7 @@
|
||||
|
||||
大家先回忆一下[77. 组合]给出的回溯法的代码：
|
||||
|
||||
```c++
|
||||
```CPP
|
||||
class Solution {
|
||||
private:
|
||||
vector<vector<int>> result; // 存放符合条件结果的集合
|
||||
@ -52,7 +52,7 @@ public:
|
||||
|
||||
在遍历的过程中有如下代码：
|
||||
|
||||
```c++
|
||||
```CPP
|
||||
for (int i = startIndex; i <= n; i++) {
|
||||
path.push_back(i);
|
||||
backtracking(n, k, i + 1);
|
||||
@ -76,7 +76,7 @@ for (int i = startIndex; i <= n; i++) {
|
||||
**如果for循环选择的起始位置之后的元素个数 已经不足 我们需要的元素个数了，那么就没有必要搜索了**。
|
||||
|
||||
注意代码中i，就是for循环里选择的起始位置。
|
||||
```c++
|
||||
```CPP
|
||||
for (int i = startIndex; i <= n; i++) {
|
||||
```
|
||||
|
||||
@ -98,13 +98,13 @@ for (int i = startIndex; i <= n; i++) {
|
||||
|
||||
所以优化之后的for循环是：
|
||||
|
||||
```c++
|
||||
```CPP
|
||||
for (int i = startIndex; i <= n - (k - path.size()) + 1; i++) // i为本次搜索的起始位置
|
||||
```
|
||||
|
||||
优化后整体代码如下：
|
||||
|
||||
```c++
|
||||
```CPP
|
||||
class Solution {
|
||||
private:
|
||||
vector<vector<int>> result;
|
||||
|
||||
@ -261,6 +261,43 @@ for (pair<string, int>target : targets[result[result.size() - 1]])
|
||||
## 其他语言版本
|
||||
|
||||
### java
|
||||
|
||||
```java
|
||||
class Solution {
|
||||
private LinkedList<String> res;
|
||||
private LinkedList<String> path = new LinkedList<>();
|
||||
|
||||
public List<String> findItinerary(List<List<String>> tickets) {
|
||||
Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));
|
||||
path.add("JFK");
|
||||
boolean[] used = new boolean[tickets.size()];
|
||||
backTracking((ArrayList) tickets, used);
|
||||
return res;
|
||||
}
|
||||
|
||||
public boolean backTracking(ArrayList<List<String>> tickets, boolean[] used) {
|
||||
if (path.size() == tickets.size() + 1) {
|
||||
res = new LinkedList(path);
|
||||
return true;
|
||||
}
|
||||
|
||||
for (int i = 0; i < tickets.size(); i++) {
|
||||
if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) {
|
||||
path.add(tickets.get(i).get(1));
|
||||
used[i] = true;
|
||||
|
||||
if (backTracking(tickets, used)) {
|
||||
return true;
|
||||
}
|
||||
|
||||
used[i] = false;
|
||||
path.removeLast();
|
||||
}
|
||||
}
|
||||
return false;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
```java
|
||||
class Solution {
|
||||
|
||||
@ -268,99 +268,102 @@ func topKFrequent(nums []int, k int) []int {
|
||||
|
||||
|
||||
|
||||
javaScript:
|
||||
JavaScript:
|
||||
```js
|
||||
/**
|
||||
* @param {number[]} nums
|
||||
* @param {number} k
|
||||
* @return {number[]}
|
||||
*/
|
||||
var topKFrequent = function(nums, k) {
|
||||
const map = new Map();
|
||||
|
||||
for(const num of nums) {
|
||||
map.set(num, (map.get(num) || 0) + 1);
|
||||
}
|
||||
|
||||
// 创建小顶堆
|
||||
const priorityQueue = new PriorityQueue((a, b) => a[1] - b[1]);
|
||||
|
||||
// entry 是一个长度为2的数组，0位置存储key，1位置存储value
|
||||
for (const entry of map.entries()) {
|
||||
priorityQueue.push(entry);
|
||||
if (priorityQueue.size() > k) {
|
||||
priorityQueue.pop();
|
||||
// js 没有堆 需要自己构造
|
||||
class Heap {
|
||||
constructor(compareFn) {
|
||||
this.compareFn = compareFn;
|
||||
this.queue = [];
|
||||
}
|
||||
}
|
||||
|
||||
const ret = [];
|
||||
// 添加
|
||||
push(item) {
|
||||
// 推入元素
|
||||
this.queue.push(item);
|
||||
|
||||
for(let i = priorityQueue.size() - 1; i >= 0; i--) {
|
||||
ret[i] = priorityQueue.pop()[0];
|
||||
}
|
||||
// 上浮
|
||||
let index = this.size() - 1; // 记录推入元素下标
|
||||
let parent = Math.floor((index - 1) / 2); // 记录父节点下标
|
||||
|
||||
return ret;
|
||||
while (parent >= 0 && this.compare(parent, index) > 0) { // 注意compare参数顺序
|
||||
[this.queue[index], this.queue[parent]] = [this.queue[parent], this.queue[index]];
|
||||
|
||||
// 更新下标
|
||||
index = parent;
|
||||
parent = Math.floor((index - 1) / 2);
|
||||
}
|
||||
}
|
||||
|
||||
// 获取堆顶元素并移除
|
||||
pop() {
|
||||
// 堆顶元素
|
||||
const out = this.queue[0];
|
||||
|
||||
// 移除堆顶元素 填入最后一个元素
|
||||
this.queue[0] = this.queue.pop();
|
||||
|
||||
// 下沉
|
||||
let index = 0; // 记录下沉元素下标
|
||||
let left = 1; // left 是左子节点下标 left + 1 则是右子节点下标
|
||||
let searchChild = this.compare(left, left + 1) > 0 ? left + 1 : left;
|
||||
|
||||
while (searchChild !== undefined && this.compare(index, searchChild) > 0) { // 注意compare参数顺序
|
||||
[this.queue[index], this.queue[searchChild]] = [this.queue[searchChild], this.queue[index]];
|
||||
|
||||
// 更新下标
|
||||
index = searchChild;
|
||||
left = 2 * index + 1;
|
||||
searchChild = this.compare(left, left + 1) > 0 ? left + 1 : left;
|
||||
}
|
||||
|
||||
return out;
|
||||
}
|
||||
|
||||
size() {
|
||||
return this.queue.length;
|
||||
}
|
||||
|
||||
// 使用传入的 compareFn 比较两个位置的元素
|
||||
compare(index1, index2) {
|
||||
// 处理下标越界问题
|
||||
if (this.queue[index1] === undefined) return 1;
|
||||
if (this.queue[index2] === undefined) return -1;
|
||||
|
||||
return this.compareFn(this.queue[index1], this.queue[index2]);
|
||||
}
|
||||
|
||||
}
|
||||
|
||||
const topKFrequent = function (nums, k) {
|
||||
const map = new Map();
|
||||
|
||||
for (const num of nums) {
|
||||
map.set(num, (map.get(num) || 0) + 1);
|
||||
}
|
||||
|
||||
// 创建小顶堆
|
||||
const heap= new Heap((a, b) => a[1] - b[1]);
|
||||
|
||||
// entry 是一个长度为2的数组，0位置存储key，1位置存储value
|
||||
for (const entry of map.entries()) {
|
||||
heap.push(entry);
|
||||
|
||||
if (heap.size() > k) {
|
||||
heap.pop();
|
||||
}
|
||||
}
|
||||
|
||||
// return heap.queue.map(e => e[0]);
|
||||
|
||||
const res = [];
|
||||
|
||||
for (let i = heap.size() - 1; i >= 0; i--) {
|
||||
res[i] = heap.pop()[0];
|
||||
}
|
||||
|
||||
return res;
|
||||
};
|
||||
|
||||
|
||||
function PriorityQueue(compareFn) {
|
||||
this.compareFn = compareFn;
|
||||
this.queue = [];
|
||||
}
|
||||
|
||||
// 添加
|
||||
PriorityQueue.prototype.push = function(item) {
|
||||
this.queue.push(item);
|
||||
let index = this.queue.length - 1;
|
||||
let parent = Math.floor((index - 1) / 2);
|
||||
// 上浮
|
||||
while(parent >= 0 && this.compare(parent, index) > 0) {
|
||||
// 交换
|
||||
[this.queue[index], this.queue[parent]] = [this.queue[parent], this.queue[index]];
|
||||
index = parent;
|
||||
parent = Math.floor((index - 1) / 2);
|
||||
}
|
||||
}
|
||||
|
||||
// 获取堆顶元素并移除
|
||||
PriorityQueue.prototype.pop = function() {
|
||||
const ret = this.queue[0];
|
||||
|
||||
// 把最后一个节点移到堆顶
|
||||
this.queue[0] = this.queue.pop();
|
||||
|
||||
let index = 0;
|
||||
// 左子节点下标，left + 1 就是右子节点下标
|
||||
let left = 1;
|
||||
let selectedChild = this.compare(left, left + 1) > 0 ? left + 1 : left;
|
||||
|
||||
// 下沉
|
||||
while(selectedChild !== undefined && this.compare(index, selectedChild) > 0) {
|
||||
// 交换
|
||||
[this.queue[index], this.queue[selectedChild]] = [this.queue[selectedChild], this.queue[index]];
|
||||
index = selectedChild;
|
||||
left = 2 * index + 1;
|
||||
selectedChild = this.compare(left, left + 1) > 0 ? left + 1 : left;
|
||||
}
|
||||
|
||||
return ret;
|
||||
}
|
||||
|
||||
PriorityQueue.prototype.size = function() {
|
||||
return this.queue.length;
|
||||
}
|
||||
|
||||
// 使用传入的 compareFn 比较两个位置的元素
|
||||
PriorityQueue.prototype.compare = function(index1, index2) {
|
||||
if (this.queue[index1] === undefined) {
|
||||
return 1;
|
||||
}
|
||||
if (this.queue[index2] === undefined) {
|
||||
return -1;
|
||||
}
|
||||
|
||||
return this.compareFn(this.queue[index1], this.queue[index2]);
|
||||
}
|
||||
```
|
||||
|
||||
TypeScript：
|
||||
|
||||
@ -118,6 +118,27 @@ class Solution {
|
||||
return res;
|
||||
}
|
||||
}
|
||||
|
||||
// 解法二
|
||||
class Solution {
|
||||
public int islandPerimeter(int[][] grid) {
|
||||
// 计算岛屿的周长
|
||||
// 方法二 : 遇到相邻的陆地总周长就-2
|
||||
int landSum = 0; // 陆地数量
|
||||
int cover = 0; // 相邻陆地数量
|
||||
for (int i = 0; i < grid.length; i++) {
|
||||
for (int j = 0; j < grid[0].length; j++) {
|
||||
if (grid[i][j] == 1) {
|
||||
landSum++;
|
||||
// 统计上面和左边的相邻陆地
|
||||
if(i - 1 >= 0 && grid[i-1][j] == 1) cover++;
|
||||
if(j - 1 >= 0 && grid[i][j-1] == 1) cover++;
|
||||
}
|
||||
}
|
||||
}
|
||||
return landSum * 4 - cover * 2;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
Python：
|
||||
|
||||
@ -12,7 +12,7 @@
|
||||
|
||||
[力扣题目链接](https://leetcode.cn/problems/reverse-string-ii/)
|
||||
|
||||
给定一个字符串 s 和一个整数 k，你需要对从字符串开头算起的每隔 2k 个字符的前 k 个字符进行反转。
|
||||
给定一个字符串 s 和一个整数 k，从字符串开头算起, 每计数至 2k 个字符，就反转这 2k 个字符中的前 k 个字符。
|
||||
|
||||
如果剩余字符少于 k 个，则将剩余字符全部反转。
|
||||
|
||||
|
||||
@ -353,8 +353,12 @@ class MyLinkedList {
|
||||
return;
|
||||
}
|
||||
size--;
|
||||
if (index == 0) {
|
||||
head = head.next;
|
||||
return;
|
||||
}
|
||||
ListNode pred = head;
|
||||
for (int i = 0; i < index; i++) {
|
||||
for (int i = 0; i < index - 1; i++) {
|
||||
pred = pred.next;
|
||||
}
|
||||
pred.next = pred.next.next;
|
||||
|
||||
Reference in New Issue
Block a user