[LeetCode] 329. Longest Increasing Path in a Matrix ☆☆☆

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.

解法：

　　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深度优先搜索DFS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：

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        
        int m = matrix.length, n = matrix[0].length;
        int res = 0;
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                res = Math.max(res, findLongestPath(matrix, dp, i, j));
            }
        }
        return res;
    }
    
    public int findLongestPath(int[][] matrix, int[][] dp, int i, int j) {
        dp[i][j] = 1;
        int[][] next = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        for (int k = 0; k < next.length; k++) {
            int ni = i + next[k][0];
            int nj = j + next[k][1];
            if (ni >= 0 && ni < dp.length && nj >= 0 && nj < dp[0].length && matrix[ni][nj] > matrix[i][j]) {
                if (dp[ni][nj] != 0) {
                    dp[i][j] = Math.max(dp[i][j], dp[ni][nj] + 1);
                } else {
                    dp[i][j] = Math.max(dp[i][j], findLongestPath(matrix, dp, ni, nj) + 1);
                }
            }
        }
        return dp[i][j];
    }
}