200. 岛屿数量

BFS或DFS

class Solution {private int[] dx = {0, 0, 1, -1};private int[] dy = {1, -1, 0, 0};int ans = 0, n, m;void dfs(char[][] grid, int x, int y){if(x < 0 || y < 0 || x >= n || y >= m || grid[x][y] == '0') return;grid[x][y] = '0';for(int i = 0; i < 4; i++){int a = x + dx[i], b = y + dy[i];dfs(grid, a, b);}}public int numIslands(char[][] grid) {n = grid.length;m = grid[0].length;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(grid[i][j] == '1'){dfs(grid, i, j);ans++;}}}return ans;}
}

198. 打家劫舍

DP问题，dp[i]表示选第i个房子时获得的最大金额，从max(dp[i-1],dp[i-2])更新答案

class Solution {public int rob(int[] nums) {int n = nums.length;for(int i = 2; i < n; i++){nums[i] = Math.max(nums[i-2], nums[Math.max(i-3, 0)]) + nums[i];}return Math.max(nums[n-1], nums[Math.max(n-2, 0)]);}
}

169. 多数元素

因为存在多数元素，多数元素的数量一定>其他元素的数量

class Solution {public int majorityElement(int[] nums) {int ans = 0, num = 0;for(int i = 0; i < nums.length; i++){if(num == 0){ans = nums[i];num++;}else if(ans == nums[i]) num++;else num--;}return ans;}
}

238. 除自身以外数组的乘积

说人话就是记一下前缀后缀

class Solution {public int[] productExceptSelf(int[] nums) {int n = nums.length;int[] ans = new int[n];for(int i = n-1; i >= 0; i--){ans[i] = (i == n-1) ? nums[i] : ans[i+1] * nums[i];//System.out.print(ans[i] + " ");}for(int i = 0; i < n; i++){int l = (i == 0) ? 1 : nums[i-1];int r = (i == n-1) ? 1 : ans[i+1];ans[i] = l * r;nums[i] *= (i == 0) ? 1 : nums[i-1];}return ans;}
}