【Lintcode】305. Longest Increasing Path in a Matrix

题目地址：

https://www.lintcode.com/problem/longest-increasing-path-in-a-matrix/description

给定一个$m$行$n$列的二维矩阵$A$，求最长严格上升路径的长度。路径的每一步允许上下左右走一步。

可以记忆化搜索。设$f[i][j]$是以$A[i][j]$为起点的最长严格上升路径的长度，则有 f [ i ] [ j ] = 1 + max ⁡ A [ i + Δ i ] [ j + Δ j ] > A [ i ] [ j ] { f [ i + Δ i ] [ j + Δ j ] } f[i][j]=1+\max_{A[i+\Delta i][j+\Delta j]>A[i][j]}\{f[i+\Delta i][j+\Delta j]\} f[i][j]=1+A[i+Δi][j+Δj]>A[i][j]max​{ f[i+Δi][j+Δj]}如果其四周都不大于$A[i][j]$则$f[i][j]=1$。为了加速可以用记忆化。代码如下：

public class Solution {
    
    
    /**
     * @param matrix: A matrix
     * @return: An integer.
     */
    public int longestIncreasingPath(int[][] matrix) {
    
    
        // Write your code here.
        int res = 0, m = matrix.length, n = matrix[0].length;
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
    
    
            for (int j = 0; j < n; j++) {
    
    
                res = Math.max(res, dfs(i, j, matrix, dp));
            }
        }
        
        return res;
    }
    
    private int dfs(int x, int y, int[][] mat, int[][] dp) {
    
    
    	// 有记忆则调取记忆
        if (dp[x][y] != 0) {
    
    
            return dp[x][y];
        }
        
        int[] d = {
    
    1, 0, -1, 0, 1};
        int res = 1;
        for (int i = 0; i < 4; i++) {
    
    
            int nextX = x + d[i], nextY = y + d[i + 1];
            if (inBound(nextX, nextY, mat) && mat[nextX][nextY] > mat[x][y]) {
    
    
                res = Math.max(res, 1 + dfs(nextX, nextY, mat, dp));
            }
        }
        
        // 存储记忆
        dp[x][y] = res;
        return res;
    }
    
    private boolean inBound(int x, int y, int[][] mat) {
    
    
        return 0 <= x && x < mat.length && 0 <= y && y < mat[0].length;
    }
}

时空复杂度$O(mn)$。