LeetCode200 岛屿数量

题目

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
 Constraints: 
 m == grid.length 
 n == grid[i].length 
 1 <= m, n <= 300 
 grid[i][j] is '0' or '1'. 

方法

深度优先遍历法

• 时间复杂度：O（）
• 空间复杂度：O（）

class Solution {
    public int numIslands(char[][] grid) {
        int row = grid.length, col = grid[0].length;
        int count = 0;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j]=='1'){
                    dfs(grid,i,j);
                    count++;
                }
            }
        }
        return count;
    }
    private void dfs(char[][] grid,int i,int j){
        int row = grid.length,col = grid[0].length;
        if(i>=row||i<0||j>=col||j<0||grid[i][j]=='0'){
            return ;
        }
        grid[i][j] = '0';
        dfs(grid,i+1,j);
        dfs(grid,i-1,j);
        dfs(grid,i,j-1);
        dfs(grid,i,j+1);
    }
}

并查集

• 时间复杂度：O（）
• 空间复杂度：O（）

标签：int,复杂度,岛屿,water,length,grid,Input,数量,LeetCode200
来源： https://www.cnblogs.com/ermiao-zy/p/15720723.html