LeetCode 200-岛屿数量

一、问题

给你一个由 ‘1’（陆地）和 ‘0’（水）组成的的二维网格，计算网格中岛屿的数量。
岛屿总是被水包围，且每座岛屿只能由水平和/或竖直方向上相邻的陆地连接形成。
此外，你可以假设该网格的四条边均被水包围。

Example:

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

二、思路

使用深度优先搜索

1. 定义一个深度优先搜索函数dfs(grid,i,j)。

• 其中grid为二维网格，i为当前位置的行数，j为当前位置的列数。具体实现：
• 将当前位置置0。从当前位置出发，判断四周是否为1，若为1，置0，递归调用dfs。

2. 遍历二维网格，出现‘1’的位点即调用dfs函数。

• 每个为‘1’的位点完成一次dfs，即形成一个岛屿。且属于该岛屿的位点都被置0，不会被再次计算。

(注意：‘0’ ‘1’的数据类型是char)

三、代码

class Solution {
public:
    void dfs(vector>& grid,int i,int j){
        int line=grid.size();
        int col=grid[0].size();

        grid[i][j]='0';

        if(j>0&&grid[i][j-1]=='1') {
            grid[i][j-1]='0';
            dfs(grid,i,j-1);
        }
        if(j0&&grid[i-1][j]=='1') {
            grid[i-1][j]='0';
            dfs(grid,i-1,j);
        }
        if(i>& grid) {
        if(grid.size()==0){
            return 0;
        }
        int island_num=0;
        int line=grid.size();
        int col=grid[0].size();
        for(int i=0;i

四、复杂度

1. 时间复杂度：m*n
2. 空间复杂度：m*n