From 39f80d5bc38868f70a85d865bba9a20fc4965cf3 Mon Sep 17 00:00:00 2001 From: Xin-Sheng-5075 <2061461846@qq.com> Date: Sun, 28 Aug 2022 19:13:24 +0800 Subject: [PATCH 1/3] Signed-off-by: Xin-Sheng-5075 <2061461846@qq.com> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加题目 200.岛屿数量 的Java版本 --- problems/0200.岛屿数量.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0200.岛屿数量.md b/problems/0200.岛屿数量.md index b88e5fd2..59c4e566 100644 --- a/problems/0200.岛屿数量.md +++ b/problems/0200.岛屿数量.md @@ -247,3 +247,35 @@ public: ## 其他语言版本 + +### Java + +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： + +```java +public int numIslands(char[][] grid) { + int res = 0; //记录找到的岛屿数量 + for(int i = 0;i < grid.length;i++){ + for(int j = 0;j < grid[0].length;j++){ + //找到“1”，res加一，同时淹没这个岛 + if(grid[i][j] == '1'){ + res++; + dfs(grid,i,j); + } + } + } + return res; +} +//使用DFS“淹没”岛屿 +public void dfs(char[][] grid, int i, int j){ + //搜索边界：索引越界或遍历到了"0" + if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0') return; + //将这块土地标记为"0" + grid[i][j] = '0'; + //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 + dfs(grid,i - 1,j); + dfs(grid,i + 1,j); + dfs(grid,i,j + 1); + dfs(grid,i,j - 1); +} +``` \ No newline at end of file From d91233d0443753b747d139e00b081fa4f4c39fb1 Mon Sep 17 00:00:00 2001 From: huaqi <2931246464@qq.com> Date: Tue, 30 Aug 2022 21:18:16 +0800 Subject: [PATCH 2/3] =?UTF-8?q?0977.=E6=9C=89=E5=BA=8F=E6=95=B0=E7=BB=84?= =?UTF-8?q?=E7=9A=84=E5=B9=B3=E6=96=B9-typescript=E7=89=88=E6=9C=AC-?= =?UTF-8?q?=E4=BC=98=E5=8C=96=E5=86=99=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0977.有序数组的平方.md | 22 ++++++++++++++-------- 1 file changed, 14 insertions(+), 8 deletions(-) diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index eb9f42b1..501b9a42 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -239,18 +239,24 @@ Typescript： ```typescript function sortedSquares(nums: number[]): number[] { - let left: number = 0, right: number = nums.length - 1; - let resArr: number[] = new Array(nums.length); - let resArrIndex: number = resArr.length - 1; + const ans: number[] = []; + let left = 0, + right = nums.length - 1; + while (left <= right) { - if (Math.abs(nums[left]) < Math.abs(nums[right])) { - resArr[resArrIndex] = nums[right--] ** 2; + // 右侧的元素不需要取绝对值，nums 为非递减排序的整数数组 + // 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 + if (Math.abs(nums[left]) > nums[right]) { + // 使用 Array.prototype.unshift() 直接在数组的首项插入当前最大值 + ans.unshift(nums[left] ** 2); + left++; } else { - resArr[resArrIndex] = nums[left++] ** 2; + ans.unshift(nums[right] ** 2); + right--; } - resArrIndex--; } - return resArr; + + return ans; }; ``` From 3f96750bece9167e25e52a61c1f52cdcdecdb8ea Mon Sep 17 00:00:00 2001 From: Wen Date: Tue, 30 Aug 2022 22:30:25 +0800 Subject: [PATCH 3/3] =?UTF-8?q?=E4=BC=98=E5=8C=96=20242.=20=E6=9C=89?= =?UTF-8?q?=E6=95=88=E7=9A=84=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D?= =?UTF-8?q?=20Java=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0242.有效的字母异位词.md | 20 +++++++++++++------- 1 file changed, 13 insertions(+), 7 deletions(-) diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 87302482..d54285eb 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -92,18 +92,24 @@ public: Java： ```java +/** + * 242. 有效的字母异位词 字典解法 + * 时间复杂度O(m+n) 空间复杂度O(1) + */ class Solution { public boolean isAnagram(String s, String t) { - int[] record = new int[26]; - for (char c : s.toCharArray()) { - record[c - 'a'] += 1; + + for (int i = 0; i < s.length(); i++) { + record[s.charAt(i) - 'a']++; } - for (char c : t.toCharArray()) { - record[c - 'a'] -= 1; + + for (int i = 0; i < t.length(); i++) { + record[t.charAt(i) - 'a']--; } - for (int i : record) { - if (i != 0) { + + for (int count: record) { + if (count != 0) { return false; } }