zoukankan      html  css  js  c++  java
  • 【leetcode】Number of Islands(middle)

    Given a 2d grid map of '1's (land) and '0's (water), count 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:

    11110
    11010
    11000
    00000

    Answer: 1

    Example 2:

    11000
    11000
    00100
    00011

    Answer: 3

    思路:与Surrounded Regions(middle)☆的思路是一样的,也可以用广度和深度优先搜索。当遇到1的时候,把区域数加1,并把与这个1相连的整片区域标记为2.

    下面的BFS用时25ms,DFS用时16ms

    int numIslands(vector<vector<char>>& grid) {
            if(grid.empty()) return 0;
            int num = 0;
            for(int i = 0; i < grid.size(); ++i)
                for(int j = 0; j < grid[0].size(); ++j)
                    if(grid[i][j] == '1')
                    {
                        num++;
                        BFS(grid, i, j);
                    }
            return num;
        }
        void BFS(vector<vector<char>>& grid, int r, int c)
        {
            queue<pair<int, int>> q;
            q.push(make_pair(r, c));
            while(!q.empty())
            {
                int i = q.front().first;
                int j = q.front().second;
                q.pop();
                if(i >= 0 && j >= 0 && i < grid.size() && j < grid[0].size() && grid[i][j] == '1')
                {
                    grid[i][j] = 2;
                    q.push(make_pair(i - 1, j));
                    q.push(make_pair(i + 1, j));
                    q.push(make_pair(i, j - 1));
                    q.push(make_pair(i, j + 1));
                }
            }
        }
        void DFS(vector<vector<char>>& grid, int r, int c)
        {
            if(r >= 0 && c >= 0 && r < grid.size() && c < grid[0].size() && grid[r][c] == '1')
            {
                grid[r][c] = 2;
                DFS(grid, r - 1, c);
                DFS(grid, r + 1, c);
                DFS(grid, r, c - 1);
                DFS(grid, r, c + 1);
            }
        }

    很奇怪的是,开始我写的时候BFS多传了一个变量就TLE了??为什么呢??

    int numIslands(vector<vector<char>>& grid) {
            if(grid.empty()) return 0;
            int num = 0;
            for(int i = 0; i < grid.size(); ++i)
            {
                for(int j = 0; j < grid[0].size(); ++j)
                {
                    if(grid[i][j] == '1')
                        BFS(grid, i, j, num);
                }
            }
            return num;
        }
    
        void BFS(vector<vector<char>>& grid, int r, int c, int &num)
        {
            num++;
            queue<pair<int, int>> q;
            q.push(make_pair(r, c));
            while(!q.empty())
            {
                int i = q.front().first;
                int j = q.front().second;
                q.pop();
                if(i >= 0 && j >= 0 && i < grid.size() && j < grid[0].size() && grid[i][j] == '1')
                {
                    grid[i][j] = num + 1;
                    q.push(make_pair(i - 1, j));
                    q.push(make_pair(i + 1, j));
                    q.push(make_pair(i, j - 1));
                    q.push(make_pair(i, j + 1));
                }
            }
        }
  • 相关阅读:
    JAVA 大数据基本操作
    C++ set 基本操作
    JVM 线上故障排查基本操作
    Git基本常用命令
    Git 入门:概念、原理、使用
    30分钟学会如何使用Shiro
    做个男人,做个成熟的男人,做个有城府的男人
    Nginx的最基本功能以及简单配置
    博客网站
    单点登录原理与简单实现
  • 原文地址:https://www.cnblogs.com/dplearning/p/4474842.html
Copyright © 2011-2022 走看看