题目:岛屿数量

给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。岛屿总是被水包围,并且每座岛屿只能由水平方向和/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。

 

输入样例1

  grid = [
      ["1","1","1","1","0"],
      ["1","1","0","1","0"],
      ["1","1","0","0","0"],
      ["0","0","0","0","0"]
      ]

输出样例1

  1

 

输入样例2

  grid = [
      ["1","1","0","0","0"],
      ["1","1","0","0","0"],
      ["0","0","1","0","0"],
      ["0","0","0","1","1"]
      ]

输出样例2

  3

 

代码:

#include<iostream>
#include<vector>
#include<deque>
using namespace std;


class Solution {
public:
    void func(vector<vector<char>>& grid)
    {
        int m = grid.size();
        int n = grid[0].size();
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (!flag[i][j] && grid[i][j]=='1')
                {
                    deque<vector<int>> deq;
                    deq.push_back({ i,j });
            flag[i][j] = 1; while (deq.size() > 0) { vector<int> tmp = deq.front(); deq.pop_front(); if (tmp[0] - 1 >= 0 && grid[tmp[0] - 1][tmp[1]] == '1' && !flag[tmp[0] - 1][tmp[1]]) { deq.push_back({ tmp[0] - 1,tmp[1] }); flag[tmp[0] - 1][tmp[1]] = 1; } if (tmp[1] - 1 >= 0 && grid[tmp[0]][tmp[1]-1] == '1' && !flag[tmp[0]][tmp[1]-1]) { deq.push_back({ tmp[0],tmp[1]-1 }); flag[tmp[0] ][tmp[1]-1] = 1; } if (tmp[0] + 1 < m && grid[tmp[0] + 1][tmp[1]] == '1' && !flag[tmp[0] + 1][tmp[1]]) { deq.push_back({ tmp[0] + 1,tmp[1] }); flag[tmp[0] + 1][tmp[1]] = 1; } if (tmp[1] + 1 < n && grid[tmp[0]][tmp[1] + 1] == '1' && !flag[tmp[0]][tmp[1] + 1]) { deq.push_back({ tmp[0],tmp[1] + 1 }); flag[tmp[0]][tmp[1] + 1] = 1; } } count++; } } } } int numIslands(vector<vector<char>>& grid) { int m = grid.size(); int n = grid[0].size(); flag = vector<vector<int>>(m, vector<int>(n, 0)); func(grid); return count; } vector<vector<int>> flag; int count = 0; };

  

posted on 2022-10-03 10:37  yc-limitless  阅读(22)  评论(0)    收藏  举报