329. Longest Increasing Path in a Matrix（第七周）

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.

解题思路：

这个题需要对每一个点进行一次DFS探索，每遇到比自己大的点则对其周围四个方向进行探索并路径长度1，若过界则返回0。则一个点的探索完毕之后将其结果保存在一个二维数组里面，然后在该数组里找出最大的一个。

#include <iostream>#include <vector>using namespace std;class Solution {private:int dfs(vector<vector<int> >&matrix, vector<vector<int> >&record, int x, int y, int jud){if(x<0 || y<0 || x>=matrix.size() || y>=matrix[0].size()) return 0; if(matrix[x][y] > jud){if(record[x][y] != 0) return record[x][y];int left  = dfs(matrix, record, x-1, y  ,matrix[x][y]) + 1;int right = dfs(matrix, record, x+1, y  ,matrix[x][y]) + 1;int up    = dfs(matrix, record, x,   y-1,matrix[x][y]) + 1;int down  = dfs(matrix, record, x,   y+1,matrix[x][y]) + 1;record[x][y] = max(left, max(right, max(up, down)));return record[x][y];}return 0; }public:    int longestIncreasingPath(vector<vector<int> >& matrix) {    int row = matrix.size();    if(row == 0) return 0;    int col = matrix[0].size();vector<vector<int> > record(row, vector<int>(col,0));int biggest = 0;for(int i = 0; i < row; i++){for(int j = 0; j< col; j++){biggest = max(biggest,dfs(matrix,record,i,j,-1));}}    return biggest;    }};