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Merge branch 'youngyangyang04:master' into master

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`@@ -247,3 +247,35 @@ public:`
`247`	`247`
`248`	`248`
`249`	`249`	`## 其他语言版本`
	`250`	`+`
	`251`	`+### Java`
	`252`	`+`
	`253`	`+下面的代码使用的是深度优先搜索 DFS 的做法。为了统计岛屿数量同时不重复记录,每当我们搜索到一个岛后,就将这个岛 "淹没" —— 将这个岛所占的地方从 "1" 改为 "0",这样就不用担心后续会重复记录这个岛屿了。而 DFS 的过程就体现在 "淹没" 这一步中。详见代码:`
	`254`	`+`
	`255`	+```java
	`256`	`+public int numIslands(char[][] grid) {`
	`257`	`+ int res = 0; //记录找到的岛屿数量`
	`258`	`+ for(int i = 0;i < grid.length;i++){`
	`259`	`+ for(int j = 0;j < grid[0].length;j++){`
	`260`	`+ //找到"1",res加一,同时淹没这个岛`
	`261`	`+ if(grid[i][j] == '1'){`
	`262`	`+ res++;`
	`263`	`+ dfs(grid,i,j);`
	`264`	`+ }`
	`265`	`+ }`
	`266`	`+ }`
	`267`	`+ return res;`
	`268`	`+}`
	`269`	`+//使用DFS"淹没"岛屿`
	`270`	`+public void dfs(char[][] grid, int i, int j){`
	`271`	`+ //搜索边界:索引越界或遍历到了"0"`
	`272`	`+ if(i < 0 \|\| i >= grid.length \|\| j < 0 \|\| j >= grid[0].length \|\| grid[i][j] == '0') return;`
	`273`	`+ //将这块土地标记为"0"`
	`274`	`+ grid[i][j] = '0';`
	`275`	`+ //根据"每座岛屿只能由水平方向或竖直方向上相邻的陆地连接形成",对上下左右的相邻顶点进行dfs`
	`276`	`+ dfs(grid,i - 1,j);`
	`277`	`+ dfs(grid,i + 1,j);`
	`278`	`+ dfs(grid,i,j + 1);`
	`279`	`+ dfs(grid,i,j - 1);`
	`280`	`+}`
	`281`	+```

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`@@ -92,18 +92,24 @@ public:`
`92`	`92`
`93`	`93`	`Java:`
`94`	`94`	```java
	`95`	`+/**`
	`96`	`+ * 242. 有效的字母异位词字典解法`
	`97`	`+ * 时间复杂度O(m+n) 空间复杂度O(1)`
	`98`	`+ */`
`95`	`99`	`class Solution {`
`96`	`100`	`public boolean isAnagram(String s, String t) {`
`97`		`-`
`98`	`101`	`int[] record = new int[26];`
`99`		`- for (char c : s.toCharArray()) {`
`100`		`- record[c - 'a'] += 1;`
	`102`	`+`
	`103`	`+ for (int i = 0; i < s.length(); i++) {`
	`104`	`+ record[s.charAt(i) - 'a']++;`
`101`	`105`	`}`
`102`		`- for (char c : t.toCharArray()) {`
`103`		`- record[c - 'a'] -= 1;`
	`106`	`+`
	`107`	`+ for (int i = 0; i < t.length(); i++) {`
	`108`	`+ record[t.charAt(i) - 'a']--;`
`104`	`109`	`}`
`105`		`- for (int i : record) {`
`106`		`- if (i != 0) {`
	`110`	`+`
	`111`	`+ for (int count: record) {`
	`112`	`+ if (count != 0) {`
`107`	`113`	`return false;`
`108`	`114`	`}`
`109`	`115`	`}`

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`@@ -239,18 +239,24 @@ Typescript:`
`239`	`239`
`240`	`240`	```typescript
`241`	`241`	`function sortedSquares(nums: number[]): number[] {`
`242`		`- let left: number = 0, right: number = nums.length - 1;`
`243`		`- let resArr: number[] = new Array(nums.length);`
`244`		`- let resArrIndex: number = resArr.length - 1;`
	`242`	`+ const ans: number[] = [];`
	`243`	`+ let left = 0,`
	`244`	`+ right = nums.length - 1;`
	`245`	`+`
`245`	`246`	`while (left <= right) {`
`246`		`- if (Math.abs(nums[left]) < Math.abs(nums[right])) {`
`247`		`- resArr[resArrIndex] = nums[right--] ** 2;`
	`247`	`+ // 右侧的元素不需要取绝对值,nums 为非递减排序的整数数组`
	`248`	`+ // 在同为负数的情况下,左侧的平方值一定大于右侧的平方值`
	`249`	`+ if (Math.abs(nums[left]) > nums[right]) {`
	`250`	`+ // 使用 Array.prototype.unshift() 直接在数组的首项插入当前最大值`
	`251`	`+ ans.unshift(nums[left] ** 2);`
	`252`	`+ left++;`
`248`	`253`	`} else {`
`249`		`- resArr[resArrIndex] = nums[left++] ** 2;`
	`254`	`+ ans.unshift(nums[right] ** 2);`
	`255`	`+ right--;`
`250`	`256`	`}`
`251`		`- resArrIndex--;`
`252`	`257`	`}`
`253`		`- return resArr;`
	`258`	`+`
	`259`	`+ return ans;`
`254`	`260`	`};`
`255`	`261`	```
`256`	`262`

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