329. Longest Increasing Path in a Matrix

1、题目描述

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].

2、解题思路：

转载：Herbert_Zero

DFS+记忆化搜索
从起点开始遍历矩阵，递归寻找其最长递增路径。
定义辅助数组record，用于记录已经搜索过的矩阵元素的数据，如record[x][y]存储了从坐标(x, y)出发的最长递增路径长度。
之后，进行深度优先搜索。逐一检查某个元素的四个方向，并继续从所有可能出现最长路径的方向上进行搜索。
当遇到record[x][y]已算出的情况，直接返回record[x][y]，减少运算量。

3、示例代码：

class Solution {

public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if (matrix.size() == 0) return 0;
        vector<int> temp(matrix[0].size(), 0);
        vector<vector<int>> record(matrix.size(), temp);
        int longest = 0;
        for (int i = 0; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[0].size(); j++) {
                longest = max(longest, DFS(matrix, record, i, j, -1));
            }
        }
        return longest;
    }
    int DFS(vector<vector<int> >& matrix, vector<vector<int> >& record, int x, int y, int lastVal) {

        if (x < 0 || y < 0 || x >= matrix.size() || y >= matrix[0].size()) return 0;
        if (matrix[x][y] > lastVal) {
            if (record[x][y] != 0) return record[x][y];
            int left = DFS(matrix, record, x - 1, y, matrix[x][y]);
            int right = DFS(matrix, record, x + 1, y, matrix[x][y]);
            int up = DFS(matrix, record, x, y + 1, matrix[x][y]);
            int bottom = DFS(matrix, record, x, y - 1, matrix[x][y]);
            record[x][y] = max(left, max(right, max(up, bottom))) + 1;
            return record[x][y];

        }

        return 0;

    }

};