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:

Input: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
Output: 4 
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
Output: 4 
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Analysis

Intuition:

以矩阵中的每一个点为出发点,如果上下左右有大于该点的则递归进行搜索,对于每一个点比较四个方向上最长的上升序列,而对全局来说则比较每个点出发的最长上升序列。Memoization可以提高DFS的性能。

Solution

DFS + DP - (12ms 55.81% AC)

class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] memo = new int[m][n];
        int max = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                max = Math.max(max, findPath(matrix, i, j, memo));
            }
        }
        return max;
    }

    int findPath(int[][] matrix, int i, int j, int[][] memo) {
        if (memo[i][j] != 0) return memo[i][j];
        int m = matrix.length;
        int n = matrix[0].length;
        int tempLength = 0;
        int[] di = new int[] {
            0,
            1,
            0,
            -1
        };
        int[] dj = new int[] {
            1,
            0,
            -1,
            0
        };

        for (int k = 0; k < 4; k++) {
            int ni = i + di[k];
            int nj = j + dj[k];
            if (ni >= 0 && nj >= 0 && ni < m && nj < n) {
                if (matrix[ni][nj] > matrix[i][j]) {
                    tempLength = Math.max(tempLength, findPath(matrix, ni, nj, memo));
                }
            }
        }
        tempLength++;
        memo[i][j] = tempLength;
        return tempLength;
    }
}

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