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

    Problem statement:

    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).

    Solution one: BFS

    This is a 2D matrix DFS and BFS problem. According to what described, the start(or end) position are four lines. the top and left lines merge into Pacific, and the right and bottom lines merge into Atlantic. The answer needs all positions that can merge into both Pacific and Atlantic.

    Push the positions in the top and left bottom boundaries into a queue, do BFS to find all positions that can merge into Pacific.

    The same for Atlantic, and find the intersection. 

    Keep two visited matrix to indicate whether the current point has already marked or not. Always pass the marked position(means could merge)(time pruning) and find candidates from unmarked.

    Time complexity is O(m * n).

    class Solution {
    public:
        vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
            if(matrix.empty()){
                return {};
            }
            int row = matrix.size();
            int col = matrix[0].size();
            queue<pair<int, int>> pacific_que;
            queue<pair<int, int>> atlantic_que;
            vector<vector<int>> in_pacific(row, vector<int>(col, 0));
            vector<vector<int>> in_atlantic(row, vector<int>(col, 0));
            vector<pair<int, int>> coordinates; 
            initialize(in_pacific, in_atlantic, pacific_que, atlantic_que, row, col);
            bfs(matrix, in_pacific, pacific_que, row, col);
            bfs(matrix, in_atlantic, atlantic_que, row, col);
            check(in_pacific, in_atlantic, coordinates, row, col);
            return coordinates;
        }
        
    private:
        void initialize(vector<vector<int>>& in_pacific, vector<vector<int>>& in_atlantic, 
                        queue<pair<int, int>>& pacific_que, queue<pair<int, int>>& atlantic_que,
                        int row, int col){
            // initialize visited matrix
            for(int j = 0; j < col; j++){
                in_pacific[0][j] = 1;
                pacific_que.push({0, j});
                in_atlantic[row - 1][j] = 1;
                atlantic_que.push({row - 1, j});
            }
            
            for(int i = 0; i < row; i++){
                in_pacific[i][0] = 1;
                pacific_que.push({i, 0});
                in_atlantic[i][col - 1] = 1;
                atlantic_que.push({i, col - 1});
            }                 
        }
    
        void bfs(vector<vector<int>>& matrix, vector<vector<int>>& visited, queue<pair<int, int>>& que, int row, int col){
            vector<pair<int,int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
            while(!que.empty()){
                pair<int, int> cur_posi = que.front();
                que.pop();
                visited[cur_posi.first][cur_posi.second] = 1;
                for(auto dir : dirs){
                    int ix = cur_posi.first + dir.first;
                    int iy = cur_posi.second + dir.second;
                    if(ix >= 0 && ix < row && iy >= 0 && iy < col && visited[ix][iy] == 0 
                    && matrix[ix][iy] >= matrix[cur_posi.first][cur_posi.second]){
                        que.push({ix, iy});
                    }
                }
            }
            return;
        }
        void check(vector<vector<int>>& pacific, vector<vector<int>> atlantic, vector<pair<int, int>>& coordinates, int row, int col){
            for(int ix = 0; ix < row; ix++){
                for(int iy = 0; iy < col; iy++){
                    if(pacific[ix][iy] & atlantic[ix][iy] == 1){
                        coordinates.push_back({ix, iy});
                    }
                }
            }
            return;
        }
    };
  • 相关阅读:
    Speech Recognize 实用类 (发现bug的朋友,请留言如何修正,供他人参考)
    由“类的成员函数”充当“回调函数”引发的问题的思考和解决方案
    装载与软件体系结构
    artoolkit video 数据转换到 IplImage*
    CvCamShift算法+原理(转)
    基于SAPI的中文语音识别的xml书写与编程
    自己根据示例代码改写的可以用于TexttoSpeech的类库
    Linux下安装erlang及rabbitmq
    jaf activation
    基于DotNetOpenAuth实现OpenID 服务提供者<shou>
  • 原文地址:https://www.cnblogs.com/wdw828/p/6855758.html
Copyright © 2011-2022 走看看