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

遍历二维数组，遇到为'1'的就递归把同一岛屿的1都置为0

• 时间复杂度：O（m*n）,m为行数，n为列数
• 空间复杂度：O（m*n）最坏情况下递归的深度

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（）