两种方式处理已经访问过的节点:一种是用visited存储已经访问过的1:另一种是通过改变原始数值的值,比如将1改成-1,这样小于等于0的都会停止. Number of Islands 用了第一种方式,Number of Distinct Islands用了第二种方式 200. Number of Islands 时间复杂度o(m*n) 1.这种写法要改变原始输入数组的值 错误版本: 条件判断顺序写错:grid[x][y] == '0' || x < 0 || x >= length || y…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by…

题目: Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded…

统计联通区域块的个数,简单dfs,请可以参考DFS框架:Leetcode 130 Surrounded Regions DFS class Solution { public: int m, n; bool is_in(int x, int y) { ) && (y < n ) && (y >= ); } void dfs(std::vector<std::vector<char>> &board, int x, int y) {…

dfs的第一题 被边界和0包围的1才是岛屿,问题就是分理出连续的1 思路是遍历数组数岛屿,dfs四个方向,遇到1后把周围连续的1置零,代表一个岛屿. /* 思路是:遍历二维数组,遇到1就把周围连续的1变成0,res+1,然后继续遍历,直到结束 周围连续1置零用的是dfs,向四个位置搜索,遇到0返回 */ public int numIslands(char[][] grid) { if (grid.length==0) return 0; int r = grid.length; int c =…