Middle-题目55：329. Longest Increasing Path in a Matrix

题目原文：
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].
题目大意：
给出一个矩阵，寻找最长的升序路径。只能沿上下左右四个方向运动。
题目分析：
从每个节点开始深搜矩阵，保持着搜索路径即可。每个dfs函数都返回从当前点出发的最大长度。
源码：（language：java）

public class Solution {
    int max = 1;
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix==null || matrix.length==0 || matrix[0].length==0) 
            return 0; 
        int[][] lenBoard = new int[matrix.length][matrix[0].length];
        int[][] dir = new int[][] { {0,0,1,-1}, {-1,1,0,0} };
        for (int i=0; i<matrix.length; ++i) {
            for (int j=0; j<matrix[0].length; ++j) {
                if (lenBoard[i][j] == 0) {
                    dfs(matrix, lenBoard, dir, i, j, Integer.MIN_VALUE);
                }
            }
        }
        return max;
    }

    private int dfs(int[][] mat, int[][] lenBoard, int[][] dir, int row, int col, int prev) {
        if (row<0 || row>=mat.length || col<0 || col>=mat[0].length || prev>=mat[row][col]) { 
            return 0;  // out of board or path stops incresing
        }
        if (lenBoard[row][col] > 0) { return lenBoard[row][col]; }  // current position already computed
        int neighborMax = 0;
        for (int i=0; i<4; ++i) {  // retrieve the length of lip from its 4 neighbors
            neighborMax = Math.max(neighborMax, dfs(mat, lenBoard, dir, row+dir[0][i], col+dir[1][i], mat[row][col]));
        }
        int localMax = 1 + neighborMax;
        lenBoard[row][col] = localMax;
        max = Math.max(max, localMax);
        return localMax;
    }
}

成绩：
16ms,beats 64.61%，众数16ms，19.26%
Cmershen的碎碎念：
本题于2016.3.11难度调整为Hard。