329. Longest Increasing Path in a Matrix(核心在于缓存遍历过程中的中间结果)

Given an integer matrix, find the length of the longest increasing path.

From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).

Example 1:

Input: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
Output: 4 
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
Output: 4 
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

class Solution {
public:
    //如果不保存计算过程中的结果，就会超时。利用dp[i][j]来保存矩阵matrix[i][j]这个位置作为起始位置的最长路径的长度。
    bool judgeValid(int x,int y,vector<vector<int>>& matrix){
        return x >= 0 && x <= matrix.size()-1 && y >=0 && y <= matrix[0].size()-1;
    }
    //x,y当前元素坐标。maxLen全局最长路径
    int dfs(vector<vector<int>>& matrix,int x,int y,vector<vector<int>> &dp){
        if(dp[x][y]) return dp[x][y];
        int maxLen = 1;
        vector<vector<int>> dirs = {{1,0},{-1,0},{0,1},{0,-1}};
        for(auto dir:dirs){
            int xx = x+dir[0];
            int yy = y+dir[1];
            if(!judgeValid(xx,yy,matrix) || matrix[xx][yy] <= matrix[x][y]) continue;
            int len = 1+dfs(matrix,xx,yy,dp);
            maxLen = max(len,maxLen);
        }
        dp[x][y] = max(maxLen,dp[x][y]);
        return dp[x][y];
    }
    //要回溯
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if(matrix.size() == 0) return 0;
        if(matrix[0].size() == 0) return 0;
        int resLen = 1;
        vector<vector<int>> dp(matrix.size(),vector<int>(matrix[0].size(),0));
        for(int i=0;i<matrix.size();i++){
            for(int j=0;j<matrix[0].size();j++){
                int len = dfs(matrix,i,j,dp);
                resLen = max(len,resLen);
            }
        }
        return resLen;
    }
};