200. Number of Islands

标签：200 return int water Number private DFSMarking grid Islands

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return 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 water.

 

Example 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

 

class Solution {

    private int n;     private int m;     public int numIslands(char[][] grid) {        int count = 0;        n = grid.length;        if (n == 0)            return 0;        m = grid[0].length;        for (int i = 0; i < n; i++) {            for (int j = 0; j < m; j++) {               if (grid[i][j] == '1') {                  DFSMarking(grid, i, j);                  count++;               }            }        }        return count;     }          private void DFSMarking(char[][] grid, int i, int j) {        if (i < 0 || j < 0 || i >= n || j>= m || grid[i][j] != '1')           return;        grid[i][j] = '0'; //这句话很关键，如果是1的节点遍历完后，会设为0，防止重复计算        DFSMarking(grid, i + 1, j);        DFSMarking(grid, i - 1, j);        DFSMarking(grid, i, j + 1);        DFSMarking(grid, i, j - 1);     } }

标签：200,return,int,water,Number,private,DFSMarking,grid,Islands
From： https://www.cnblogs.com/MarkLeeBYR/p/16921021.html