Merge branch 'youngyangyang04:master' into master

This commit is contained in:
Yukun J
2022-08-30 21:47:43 -04:00
committed by GitHub
9 changed files with 247 additions and 162 deletions

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@ -263,21 +263,38 @@ public:
Java Java
```java ```java
public int[] twoSum(int[] nums, int target) { // 双指针 动态规划
int[] res = new int[2]; class Solution {
if(nums == null || nums.length == 0){ public String longestPalindrome(String s) {
return res; 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<Integer, Integer> map = new HashMap<>();
for(int i = 0; i < nums.length; i++){ for (int L = 2; L <= s.length(); L++) {
int temp = target - nums[i]; for (int i = 0; i < s.length(); i++) {
if(map.containsKey(temp)){ int j = i + L - 1;
res[1] = i; if (j >= s.length()) break;
res[0] = map.get(temp); 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];
} }
map.put(nums[i], i);
} }
return res; if (palindrome[i][j] && j - i + 1 > length) {
length = j - i + 1;
index = i;
}
}
}
return s.substring(index, index + length);
}
} }
``` ```

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@ -222,56 +222,6 @@ public:
## 其他语言补充 ## 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
```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
```Go ```Go
@ -396,6 +395,8 @@ func isValid(n, row, col int, chessboard [][]string) bool {
return true return true
} }
``` ```
### Javascript ### Javascript
```Javascript ```Javascript
var solveNQueens = function(n) { var solveNQueens = function(n) {

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@ -188,11 +188,13 @@ JavaScript:
```javascript ```javascript
var climbStairs = function(n) { var climbStairs = function(n) {
const dp = new Array(n + 1).fill(0); const dp = new Array(n + 1).fill(0);
const weight = [1,2]; const m = 2;
dp[0] = 1; dp[0] = 1;
for(let i = 0; i <= n; i++){ //先遍历背包 for(let i = 1; i <= n; i++){
for(let j = 0; j < weight.length; j++){ // 再遍历物品 for(let j = 1; j <= m; j++){
if(i >= weight[j]) dp[i] += dp[i-weight[j]]; if(i >= j) {
dp[i] += dp[i - j];
}
} }
} }
return dp[n]; return dp[n];

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@ -20,7 +20,7 @@
大家先回忆一下[77. 组合]给出的回溯法的代码: 大家先回忆一下[77. 组合]给出的回溯法的代码:
```c++ ```CPP
class Solution { class Solution {
private: private:
vector<vector<int>> result; // 存放符合条件结果的集合 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: 在遍历的过程中有如下代码:
```c++ ```CPP
for (int i = startIndex; i <= n; i++) { for (int i = startIndex; i <= n; i++) {
path.push_back(i); path.push_back(i);
backtracking(n, k, i + 1); 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** **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。 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++) { 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 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 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: 优化后整体代码如下:
```c++ ```CPP
class Solution { class Solution {
private: private:
vector<vector<int>> result; vector<vector<int>> result;

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@ -262,6 +262,43 @@ for (pair<string, int>target : targets[result[result.size() - 1]])
### java ### 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 ```java
class Solution { class Solution {
private Deque<String> res; private Deque<String> res;

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@ -268,14 +268,74 @@ func topKFrequent(nums []int, k int) []int {
javaScript: JavaScript:
```js ```js
/** // js 没有堆 需要自己构造
* @param {number[]} nums class Heap {
* @param {number} k constructor(compareFn) {
* @return {number[]} this.compareFn = compareFn;
*/ this.queue = [];
var topKFrequent = function(nums, k) { }
// 添加
push(item) {
// 推入元素
this.queue.push(item);
// 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
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(); const map = new Map();
for (const num of nums) { for (const num of nums) {
@ -283,84 +343,27 @@ var topKFrequent = function(nums, k) {
} }
// 创建小顶堆 // 创建小顶堆
const priorityQueue = new PriorityQueue((a, b) => a[1] - b[1]); 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key1位置存储value // entry 是一个长度为2的数组0位置存储key1位置存储value
for (const entry of map.entries()) { for (const entry of map.entries()) {
priorityQueue.push(entry); heap.push(entry);
if (priorityQueue.size() > k) {
priorityQueue.pop(); if (heap.size() > k) {
heap.pop();
} }
} }
const ret = []; // return heap.queue.map(e => e[0]);
for(let i = priorityQueue.size() - 1; i >= 0; i--) { const res = [];
ret[i] = priorityQueue.pop()[0];
for (let i = heap.size() - 1; i >= 0; i--) {
res[i] = heap.pop()[0];
} }
return ret; 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 TypeScript

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@ -118,6 +118,27 @@ class Solution {
return res; 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 Python

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@ -12,7 +12,7 @@
[力扣题目链接](https://leetcode.cn/problems/reverse-string-ii/) [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。 如果剩余字符少于 k 个,则将剩余字符全部反转。

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@ -353,8 +353,12 @@ class MyLinkedList {
return; return;
} }
size--; size--;
if (index == 0) {
head = head.next;
return;
}
ListNode pred = head; ListNode pred = head;
for (int i = 0; i < index; i++) { for (int i = 0; i < index - 1; i++) {
pred = pred.next; pred = pred.next;
} }
pred.next = pred.next.next; pred.next = pred.next.next;