Longest Increasing Path in a Matrix

LeetCode #329
Interview

Description:

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:

Example 1:

nums = [
  [9,9,4],
  [6,6,8],
  [2,1,1]
]
Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

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

Idea:

Note: 其实就是2D的DP。

核心思想是DFS+DP，DP用来存已经计算了的值，DFS一路到底。

以下为两种解法，prefer第一种，更清楚点，差别是对于不符合条件的情况是先搜再返回0还是不搜。

Code:

方法一，expand的话不符合条件不继续，如果没有expand的话，该点设为1（不能是0）。

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if(matrix.empty()) return 0;

        int m=matrix.size();
        int n=matrix[0].size();

        vector<vector<int>> DP(m, vector<int>(n, 0));

        int max_path=0;

        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(DP[i][j]==0){
                    max_path = max(max_path, DFS(matrix, i, j, DP));
                }
            }
        }

        return max_path;
    }

    int DFS(const vector<vector<int>>& matrix, int i, int j, vector<vector<int>>& DP){
        int m=matrix.size();
        int n=matrix[0].size();

        if(DP[i][j]!=0) return DP[i][j];   

        vector<pair<int, int>> move={{-1,0}, {0,1}, {1,0}, {0,-1}};

        int val=0;

        for(auto e: move){
            int i_next = i+e.first;
            int j_next = j+e.second;

            if(i_next>=0 && i_next<m && j_next>=0 && j_next<n && matrix[i][j]<matrix[i_next][j_next])
                val = max(val, DFS(matrix, i_next, j_next, DP));
        }
        DP[i][j] = val+1;

        return DP[i][j];
    }
};

方法二，不管，直接搜，function一开始判断情况，不满足条件返回0.但是第一次call的时候没有上一次的值，伪造一个。

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if(matrix.empty()) return 0;

        int m=matrix.size();
        int n=matrix[0].size();

        vector<vector<int>> DP(m, vector<int>(n, 0));

        int max_path=0;

        for(int i=0; i<m; ++i){
            for(int j=0; j<n; ++j){
                if(DP[i][j]==0){
                    max_path = max(max_path, DFS(matrix, matrix[i][j]-1, i, j, DP)); // Assume the pre_val is matrix[i][j]-1, otherwise always return 0;
                }
            }
        }

        return max_path;
    }

    int DFS(const vector<vector<int>>& matrix, int pre_val, int i, int j, vector<vector<int>>& DP){
        int m=matrix.size();
        int n=matrix[0].size();

        if(i<0 || i>=m || j<0 || j>=n) return 0;

        if(matrix[i][j]<=pre_val) return 0;

        if(DP[i][j]!=0) return DP[i][j];   

        vector<pair<int, int>> move={{-1,0}, {0,1}, {1,0}, {0,-1}};

        int val=0;

        for(auto e: move){
            int i_next = i+e.first;
            int j_next = j+e.second;
            val = max(val, DFS(matrix, matrix[i][j], i_next, j_next, DP));
        }
        DP[i][j] = val+1;

        return DP[i][j];
    }
};