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Commit a9d679b

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Merge branch 'youngyangyang04:master' into master
2 parents 26cdabc + 07e0016 commit a9d679b

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-5
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+77
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‎problems/0027.移除元素.md‎

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```java
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class Solution {
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public int removeElement(int[] nums, int val) {
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// 快慢指针
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int fastIndex = 0;
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int slowIndex;
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for (slowIndex = 0; fastIndex < nums.length; fastIndex++) {
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int slowIndex = 0;
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for (int fastIndex = 0; fastIndex < nums.length; fastIndex++) {
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if (nums[fastIndex] != val) {
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nums[slowIndex] = nums[fastIndex];
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slowIndex++;
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}
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}
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return slowIndex;
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}
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}
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```

‎problems/0150.逆波兰表达式求值.md‎

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# 思路
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《代码随想录》算法视频公开课:[栈的最后表演! | LeetCode:150. 逆波兰表达式求值](https://www.bilibili.com/video/BV1kd4y1o7on),相信结合视频在看本篇题解,更有助于大家对本题的理解。
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在上一篇文章中[1047.删除字符串中的所有相邻重复项](https://programmercarl.com/1047.删除字符串中的所有相邻重复项.html)提到了 递归就是用栈来实现的。
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所以**栈与递归之间在某种程度上是可以转换的!** 这一点我们在后续讲解二叉树的时候,会更详细的讲解到。

‎problems/0239.滑动窗口最大值.md‎

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# 思路
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《代码随想录》算法视频公开课:[单调队列正式登场!| LeetCode:239. 滑动窗口最大值](https://www.bilibili.com/video/BV1XS4y1p7qj),相信结合视频在看本篇题解,更有助于大家对本题的理解。
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这是使用单调队列的经典题目。
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难点是如何求一个区间里的最大值呢? (这好像是废话),暴力一下不就得了。

‎problems/0347.前K个高频元素.md‎

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# 思路
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《代码随想录》算法视频公开课:[优先级队列正式登场!大顶堆、小顶堆该怎么用?| LeetCode:347.前 K 个高频元素](https://www.bilibili.com/video/BV1Xg41167Lz),相信结合视频在看本篇题解,更有助于大家对本题的理解。
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<p align="center">
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<iframe src="//player.bilibili.com/player.html?aid=514643371&bvid=BV1Xg41167Lz&cid=808260290&page=1" scrolling="no" border="0" frameborder="no" framespacing="0" allowfullscreen="true" width=750 height=500></iframe>
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</p>
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这道题目主要涉及到如下三块内容:
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1. 要统计元素出现频率
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2. 对频率排序

‎problems/0695.岛屿的最大面积.md‎

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# 思路
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这道题目也是 dfs bfs基础类题目。
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## DFS
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很多同学,写dfs其实也是凭感觉来,有的时候dfs函数中写终止条件才能过,有的时候 dfs函数不写终止添加也能过!
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这里其实涉及到dfs的两种写法,
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以下代码使用dfs实现,如果对dfs不太了解的话,建议先看这篇题解:[797.所有可能的路径](https://leetcode.cn/problems/all-paths-from-source-to-target/solution/by-carlsun-2-66pf/),
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写法一,dfs只处理下一个节点
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```CPP
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class Solution {
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}
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};
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```
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以上两种写法的区别,我在题解: [DFS,BDF 你没注意的细节都给你列出来了!LeetCode:200. 岛屿数量](https://leetcode.cn/problems/number-of-islands/solution/by-carlsun-2-n72a/)做了详细介绍。
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## BFS
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```CPP
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class Solution {
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private:
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int count;
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int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
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void bfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
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queue<int> que;
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que.push(x);
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que.push(y);
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visited[x][y] = true; // 加入队列就意味节点是陆地可到达的点
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count++;
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while(!que.empty()) {
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int xx = que.front();que.pop();
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int yy = que.front();que.pop();
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for (int i = 0 ;i < 4; i++) {
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int nextx = xx + dir[i][0];
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int nexty = yy + dir[i][1];
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if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界
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if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 节点没有被访问过且是陆地
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visited[nextx][nexty] = true;
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count++;
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que.push(nextx);
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que.push(nexty);
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}
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}
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}
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}
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public:
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int maxAreaOfIsland(vector<vector<int>>& grid) {
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int n = grid.size(), m = grid[0].size();
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vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
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int result = 0;
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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if (!visited[i][j] && grid[i][j] == 1) {
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count = 0;
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bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
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result = max(result, count);
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}
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}
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}
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return result;
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}
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};
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```

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