天津网站建设培训课件头条今日头条新闻

代码随想录图论 第二天 | 695. 岛屿的最大面积 1020. 飞地的数量

一、695. 岛屿的最大面积

题目链接：https://leetcode.cn/problems/max-area-of-island/
思路：典型的遍历模板题，我采用深度优先，每块岛屿递归遍历的时候计数，递归完比较大小记录最大值。

class Solution {int max = 0, k = 0;public int maxAreaOfIsland(int[][] grid) {for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j] == 1) {dfs(grid, i, j);max = Math.max(max, k);k = 0;}}}return max;}void dfs(int[][] grid, int x, int y) {if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] != 1) {return;}k++;grid[x][y] = 0;dfs(grid, x, y-1);dfs(grid, x, y+1);dfs(grid, x-1, y);dfs(grid, x+1, y);}
}

二、1020. 飞地的数量

题目链接：https://leetcode.cn/problems/number-of-enclaves/description/
思路：求飞地的数量其实就是求不与边框相接的地块数量，那么可以留一个标识位flag，递归中发现不是飞地标记一下，该次递归记得数就不累加了。

class Solution {// true表示没有接触边界boolean flag = true;int num = 0;public int numEnclaves(int[][] grid) {int sum = 0;for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j] == 1) {dfs(grid, i, j);if (flag) {sum += num;}else {flag = true;}num = 0;}}}return sum;}void dfs(int[][] grid, int x, int y) {if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] != 1) {return;}if (x == 0 || y == 0 || x == grid.length-1 || y == grid[0].length-1) {flag = false;}num++;grid[x][y] = 0;dfs(grid, x, y-1);dfs(grid, x, y+1);dfs(grid, x-1, y);dfs(grid, x+1, y);}
}