zoukankan      html  css  js  c++  java
  • 417. Pacific Atlantic Water Flow

    Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.

    Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.

    Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.

    Note:

    1. The order of returned grid coordinates does not matter.
    2. Both m and n are less than 150.

    Example:

    Given the following 5x5 matrix:
    
      Pacific ~   ~   ~   ~   ~ 
           ~  1   2   2   3  (5) *
           ~  3   2   3  (4) (4) *
           ~  2   4  (5)  3   1  *
           ~ (6) (7)  1   4   5  *
           ~ (5)  1   1   2   4  *
              *   *   *   *   * Atlantic
    
    Return:
    
    [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
    

     解法1:DFS,遍历每一个元素,找到路径,复杂度过高,有些case超时

    class Solution {
    public:
        vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
            vector<vector<int>>res;
            int m = matrix.size();
            if (m < 1)return res;
            int n = matrix[0].size();
            if (n < 1)return res;
            
            for (int i = 0; i < m; ++i)
            {
                for(int j = 0; j < n; ++j)
                {
                    bool f1 = false, f2 = false;
                    vector<vector<bool>> marked(m, vector<bool>(n, false));
                    
                    dfs(matrix, i, j, f1, f2, marked);
                    
                    if (f1 && f2)res.push_back(vector<int>{i,j});
                }
            }
            return res;
        }
        
        void dfs(vector<vector<int>>& matrix, int i, int j, bool& flag1, bool& flag2, vector<vector<bool>>& marked)
        {
            if ((flag1 && flag2) || marked[i][j])return;
            int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
            marked[i][j] = true;
            for (auto d: dir)
            {
                if (i+d[0] < 0 || j+d[1] < 0) // 当前格子已经第一排或第一列了
                {
                    flag1 = true;
                    continue;
                }
                if (i+d[0] >= matrix.size() || j+d[1] >= matrix[0].size()) // 当前格子已经最后一排或者最后一列了
                {
                    flag2 = true;
                    continue;
                }
                // 下一个方格比当前的低,并且之前没有走过
                if(matrix[i+d[0]][j+d[1]] <= matrix[i][j] && !marked[i+d[0]][j+d[1]])
                    dfs(matrix, i+d[0], j+d[1], flag1, flag2, marked);
                if (flag1 && flag2)return;
            }
        }
    };

    解法2:DFS从太平洋和大西洋往内部遍历

    class Solution {
    public:
        vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
            vector<vector<int>>res;
            int m = matrix.size();
            if (m < 1)return res;
            int n = matrix[0].size();
            if (n < 1)return res;
            vector<vector<bool>> marked1(m, vector<bool>(n, false));
            vector<vector<bool>> marked2(m, vector<bool>(n, false));
            for (int i = 0;i < m; ++i)
            {
                dfs(matrix, i, 0, marked1);
                dfs(matrix, i, n-1, marked2);
            }
            for (int i = 0;i < n;++i)
            {
                dfs(matrix, 0, i, marked1);
                dfs(matrix, m-1, i, marked2);
            }
            
            for (int i = 0; i < m; ++i)
            {
                for(int j = 0; j < n; ++j)
                {
                    if (marked1[i][j] && marked2[i][j])res.push_back(vector<int>{i,j});
                }
            }
            return res;
        }
        
        void dfs(vector<vector<int>>& matrix, int i, int j, vector<vector<bool>>& marked)
        {
            if (marked[i][j])return; 
            int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
            marked[i][j] = true;
            for (auto d: dir)
            {
                int i_next = i + d[0];
                int j_next = j + d[1];
    
                if(i_next < 0 || i_next >= matrix.size() || j_next < 0 || j_next >= matrix[0].size() || 
                  matrix[i][j] > matrix[i_next][j_next])
                    continue;
                dfs(matrix, i+d[0], j+d[1], marked);
            }
        }
    };
  • 相关阅读:
    Asp.net 程序连接orcle如果在安装 32 位 Oracle 客户端组件的情况下以 64 位模式运行,
    Navicat 远程连接 Oracle11g 数据库报错 No listener 的问题
    springMVC中@DateTimeFormat 失效的处理
    单例设计模式中懒汉式线程安全的处理
    ajax同步请求
    vue.js在标签属性中拼接字符串
    vue.js进行遍历
    html页面之间的传值
    URL编码和解码的一个小问题(JS方法和JAVA方法)
    SolrJ的配置及使用
  • 原文地址:https://www.cnblogs.com/qiang-wei/p/12359484.html
Copyright © 2011-2022 走看看