329. Longest Increasing Path in a Matrix (Hard)

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.

Solution: DFS

这道题给我们一个二维数组，让我们求矩阵中最长的递增路径，规定我们只能上下左右行走，不能走斜线或者是超过了边界。那么这道题的解法要用递归和DP来解，用DP的原因是为了提高效率，避免重复运算。我们需要维护一个二维动态数组dp，其中dp[i][j]表示数组中以(i,j)为起点的最长递增路径的长度，初始将dp数组都赋为0，当我们用递归调用时，遇到某个位置(x, y), 如果dp[x][y]不为0的话，我们直接返回dp[x][y]即可，不需要重复计算。我们需要以数组中每个位置都为起点调用递归来做，比较找出最大值。在以一个位置为起点用DFS搜索时，对其四个相邻位置进行判断，如果相邻位置的值大于上一个位置，则对相邻位置继续调用递归，并更新一个最大值，搜素完成后返回即可，参见代码如下：

same value cannot appear in the same path!

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n = matrix.size(); if (n == 0) return 0;
        int m = matrix[0].size(); if (m == 0) return 0;
        int res = 1;
        vector<vector<int>> dp(n, vector<int>(m, 0));

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                res = max(res, dfs(matrix, dp, i, j));
            }
        }
        return res;
    }

    int dfs(vector<vector<int>>& matrix, vector<vector<int>>& dp, int i, int j) {
        if (dp[i][j]) return dp[i][j]; // visited
        static vector<pair<int, int>> dirs = {{0,-1}, {-1,0}, {0, 1}, {1,0}};
        int n = matrix.size(), m = matrix[0].size(), maxlen = 1;

        for (auto a: dirs) {
            int x = i+a.first, y = j+a.second;
            // check out of bound and whether directed edge exists (must be <=!)
            if (x < 0 || x >= n || y < 0 || y >= m || matrix[x][y] <= matrix[i][j]) continue;
            int len = 1+dfs(matrix, dp, x, y);
            maxlen = max(maxlen, len);
        }

        dp[i][j] = maxlen;
        return maxlen;
    }

};