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

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programmercarl
2024-06-24 17:13:27 +08:00
parent 26a5b0cc21 05bc2cef6f
commit bcbd631dc9
7 changed files with 185 additions and 31 deletions
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3
problems/0101.对称二叉树.md
View File

@ -224,8 +224,8 @@ public:
st.push(root->left);
st.push(root->right);
while (!st.empty()) {
TreeNode* leftNode = st.top(); st.pop();
TreeNode* rightNode = st.top(); st.pop();
TreeNode* leftNode = st.top(); st.pop();
if (!leftNode && !rightNode) {
continue;
}
@ -950,3 +950,4 @@ public bool IsSymmetric(TreeNode root)
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
</a>

25
problems/0242.有效的字母异位词.md
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@ -383,6 +383,31 @@ object Solution {
}
```
### C
```c
bool isAnagram(char* s, char* t) {
int len1 = strlen(s), len2 = strlen(t);
if (len1 != len2) {
return false;
}
int map1[26] = {0}, map2[26] = {0};
for (int i = 0; i < len1; i++) {
map1[s[i] - 'a'] += 1;
map2[t[i] - 'a'] += 1;
}
for (int i = 0; i < 26; i++) {
if (map1[i] != map2[i]) {
return false;
}
}
return true;
}
```
## 相关题目
* [383.赎金信](https://programmercarl.com/0383.%E8%B5%8E%E9%87%91%E4%BF%A1.html)

36
problems/0454.四数相加II.md
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@ -185,22 +185,28 @@ class Solution:
### Go：
```go
func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
m := make(map[int]int) //key:a+b的数值，value:a+b数值出现的次数
count := 0
// 遍历nums1和nums2数组，统计两个数组元素之和，和出现的次数，放到map中
func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
m := make(map[int]int)
count := 0
// 构建nums1和nums2的和的map
for _, v1 := range nums1 {
for _, v2 := range nums2 {
m[v1+v2]++
}
}
// 遍历nums3和nums4数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来
for _, v3 := range nums3 {
for _, v4 := range nums4 {
count += m[-v3-v4]
}
}
return count
for _, v2 := range nums2 {
m[v1+v2]++
}
}
// 遍历nums3和nums4，检查-c-d是否在map中
for _, v3 := range nums3 {
for _, v4 := range nums4 {
sum := -v3 - v4
if countVal, ok := m[sum]; ok {
count += countVal
}
}
}
return count
}
```

20
problems/0704.二分查找.md
View File

@ -174,13 +174,17 @@ class Solution {
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + ((right - left) >> 1);
if (nums[mid] == target)
if (nums[mid] == target) {
return mid;
else if (nums[mid] < target)
}
else if (nums[mid] < target) {
left = mid + 1;
else if (nums[mid] > target)
}
else { // nums[mid] > target
right = mid - 1;
}
}
// 未找到目标值
return -1;
}
}
@ -194,13 +198,17 @@ class Solution {
int left = 0, right = nums.length;
while (left < right) {
int mid = left + ((right - left) >> 1);
if (nums[mid] == target)
if (nums[mid] == target) {
return mid;
else if (nums[mid] < target)
}
else if (nums[mid] < target) {
left = mid + 1;
else if (nums[mid] > target)
}
else { // nums[mid] > target
right = mid;
}
}
// 未找到目标值
return -1;
}
}

58
problems/1020.飞地的数量.md
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@ -605,6 +605,63 @@ func bfs(grid [][]int, i, j int) {
}
```
### JavaScript
```js
/**
* @param {number[][]} grid
* @return {number}
*/
var numEnclaves = function (grid) {
let row = grid.length;
let col = grid[0].length;
let count = 0;
// Check the first and last row, if there is a 1, then change all the connected 1s to 0 and don't count them.
for (let j = 0; j < col; j++) {
if (grid[0][j] === 1) {
dfs(0, j, false);
}
if (grid[row - 1][j] === 1) {
dfs(row - 1, j, false);
}
}
// Check the first and last column, if there is a 1, then change all the connected 1s to 0 and don't count them.
for (let i = 0; i < row; i++) {
if (grid[i][0] === 1) {
dfs(i, 0, false);
}
if (grid[i][col - 1] === 1) {
dfs(i, col - 1, false);
}
}
// Check the rest of the grid, if there is a 1, then change all the connected 1s to 0 and count them.
for (let i = 1; i < row - 1; i++) {
for (let j = 1; j < col - 1; j++) {
dfs(i, j, true);
}
}
function dfs(i, j, isCounting) {
let condition = i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] === 0;
if (condition) return;
if (isCounting) count++;
grid[i][j] = 0;
dfs(i - 1, j, isCounting);
dfs(i + 1, j, isCounting);
dfs(i, j - 1, isCounting);
dfs(i, j + 1, isCounting);
}
return count;
};
```
### Rust
dfs:
@ -700,3 +757,4 @@ impl Solution {
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
</a>

18
problems/1207.独一无二的出现次数.md
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@ -40,9 +40,9 @@
回归本题，**本题强调了-1000 <= arr[i] <= 1000**，那么就可以用数组来做哈希，arr[i]作为哈希表（数组）的下标，那么arr[i]可以是负数，怎么办？负数不能做数组下标。
**此时可以定义一个2000大小的数组，例如int count[2002];**，统计的时候，将arr[i]统一加1000，这样就可以统计arr[i]的出现频率了。
**此时可以定义一个2001大小的数组，例如int count[2001];**，统计的时候，将arr[i]统一加1000，这样就可以统计arr[i]的出现频率了。
题目中要求的是是否有相同的频率出现，那么需要再定义一个哈希表（数组）用来记录频率是否重复出现过，bool fre[1002]; 定义布尔类型的就可以了，**因为题目中强调1 <= arr.length <= 1000，所以哈希表大小为1000就可以了**。
题目中要求的是是否有相同的频率出现，那么需要再定义一个哈希表（数组）用来记录频率是否重复出现过，bool fre[1001]; 定义布尔类型的就可以了，**因为题目中强调1 <= arr.length <= 1000，所以哈希表大小为1000就可以了**。
如图所示：
@ -55,11 +55,11 @@ C++代码如下：
class Solution {
public:
bool uniqueOccurrences(vector<int>& arr) {
int count[2002] = {0}; // 统计数字出现的频率
int count[2001] = {0}; // 统计数字出现的频率
for (int i = 0; i < arr.size(); i++) {
count[arr[i] + 1000]++;
}
bool fre[1002] = {false}; // 看相同频率是否重复出现
bool fre[1001] = {false}; // 看相同频率是否重复出现
for (int i = 0; i <= 2000; i++) {
if (count[i]) {
if (fre[count[i]] == false) fre[count[i]] = true;
@ -78,7 +78,7 @@ public:
```java
class Solution {
public boolean uniqueOccurrences(int[] arr) {
int[] count = new int[2002];
int[] count = new int[2001];
for (int i = 0; i < arr.length; i++) {
count[arr[i] + 1000]++; // 防止负数作为下标
}
@ -103,10 +103,10 @@ class Solution {
# 方法 1: 数组在哈西法的应用
class Solution:
def uniqueOccurrences(self, arr: List[int]) -> bool:
count = [0] * 2002
count = [0] * 2001
for i in range(len(arr)):
count[arr[i] + 1000] += 1 # 防止负数作为下标
freq = [False] * 1002 # 标记相同频率是否重复出现
freq = [False] * 1001 # 标记相同频率是否重复出现
for i in range(2001):
if count[i] > 0:
if freq[count[i]] == False:
@ -139,12 +139,12 @@ class Solution:
``` javascript
// 方法一：使用数组记录元素出现次数
var uniqueOccurrences = function(arr) {
const count = new Array(2002).fill(0);// -1000 <= arr[i] <= 1000
const count = new Array(2001).fill(0);// -1000 <= arr[i] <= 1000
for(let i = 0; i < arr.length; i++){
count[arr[i] + 1000]++;// 防止负数作为下标
}
// 标记相同频率是否重复出现
const fre = new Array(1002).fill(false);// 1 <= arr.length <= 1000
const fre = new Array(1001).fill(false);// 1 <= arr.length <= 1000
for(let i = 0; i <= 2000; i++){
if(count[i] > 0){//有i出现过
if(fre[count[i]] === false) fre[count[i]] = true;//之前未出现过，标记为出现

56
problems/二叉树的递归遍历.md
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@ -671,6 +671,62 @@ public void Traversal(TreeNode cur, IList<int> res)
}
```
### PHP
```php
// 144.前序遍历
function preorderTraversal($root) {
$output = [];
$this->traversal($root, $output);
return $output;
}
function traversal($root, array &$output) {
if ($root->val === null) {
return;
}
$output[] = $root->val;
$this->traversal($root->left, $output);
$this->traversal($root->right, $output);
}
```
```php
// 94.中序遍历
function inorderTraversal($root) {
$output = [];
$this->traversal($root, $output);
return $output;
}
function traversal($root, array &$output) {
if ($root->val === null) {
return;
}
$this->traversal($root->left, $output);
$output[] = $root->val;
$this->traversal($root->right, $output);
}
```
```php
// 145.后序遍历
function postorderTraversal($root) {
$output = [];
$this->traversal($root, $output);
return $output;
}
function traversal($root, array &$output) {
if ($root->val === null) {
return;
}
$this->traversal($root->left, $output);
$this->traversal($root->right, $output);
$output[] = $root->val;
}
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">