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[Google] LeetCode 329 Longest Increasing Path in a Matrix 记忆化搜索

2022-09-04 06:30:08 作者：互联网

Given an m x n integers matrix, return the length of the longest increasing path in matrix.

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

Solution

很容易的一个做法就是枚举每个点，然后对每个点进行 \(DFS\). 但是会 \(TLE\).

考虑记忆化搜索，维护一个 \(dp\) 数组：

\[dp[x][y]=\max(dp[x][y], 1+dfs(nx,ny)) \]

初始化 \(dp\) 我们用 \(-1\).

点击查看代码

class Solution {
private:
    int dir[4][2]={
        1,0,
        0,1,
        -1,0,
        0,-1
    };
    
    bool check(int x, int y,int r, int c){
        if(x<0||y<0||x>=r||y>=c) return false;
        return true;
    }
    int ans=0;
    
    
    int dfs(int x, int y, int r, int c, vector<vector<int>>& m, vector<vector<int>>&dp){
        if(dp[x][y]!=-1)return dp[x][y];
        dp[x][y]=1;
        for(int i=0;i<4;i++){
            int nx=x+dir[i][0], ny=y+dir[i][1];
            if(check(nx,ny,r,c)&& m[nx][ny]>m[x][y]){
                dp[x][y]=max(dp[x][y],1+dfs(nx,ny,r,c,m,dp));
            }
        }
        return dp[x][y];
    }
    
    
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int r = matrix.size(), c = matrix[0].size();
        vector<vector<int>> dp(r, vector<int>(c,-1));
        
        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
                ans = max(ans, dfs(i,j,r,c,matrix,dp));
            }
        }
        return ans;
    }
};

标签：Google,return,Matrix,int,move,vector,Increasing,dp,matrix
来源： https://www.cnblogs.com/xinyu04/p/16654189.html