[leetcode]329. Longest Increasing Path in a Matrix

题目链接：https://leetcode.com/problems/longest-increasing-path-in-a-matrix/#/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 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.

class Solution {public:    int DFS(vector<vector<int>>& matrix, int x, int y, int val, vector<vector<int>>& hash)    {        if(x < 0 || x >= matrix.size() || y <0 || y >= matrix[0].size())            return 0;        if(matrix[x][y] > val)        {            if(hash[x][y] != 0) return hash[x][y]; //if this path has been searched            int a = DFS(matrix, x-1, y,matrix[x][y], hash) + 1;            int b = DFS(matrix, x+1, y,matrix[x][y], hash) + 1;            int c = DFS(matrix, x, y-1,matrix[x][y], hash) + 1;            int d = DFS(matrix, x, y+1,matrix[x][y], hash) + 1;            hash[x][y] = max(a, max(b,max(c, d)));            return hash[x][y];        }        return 0;    }    int longestIncreasingPath(vector<vector<int>>& matrix) {        if(matrix.empty()) return 0;        int Max = 0;        vector<vector<int>> hash(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++)                Max = max(DFS(matrix, i, j, INT32_MIN, hash), Max);        return Max;    }};