[leetcode]longest increasing path in a matrix

题目描述

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.

代码

public class  {
    
    public static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

    public int longestIncreasingPath(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int row = matrix.length, col = matrix[0].length;
        int max = 1;
        int[][] cache = new int[row][col];
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                int len = dfs(matrix, i, j, row, col, cache);
                max = Math.max(max, len);
            }
        }

        return max;
    }

    //深度搜索从[i][j]出发的最长递增路径的长度
    //cache缓存，即备忘录
    public int dfs(int[][] matrix, int i, int j, int row, int col, int[][] cache){
        //已经搜索过，直接返回
        if(cache[i][j] != 0) return cache[i][j];
        int max = 1;
        //四个方向搜索
        for(int[] dir : dirs){
            int x = i + dir[0], y = j + dir[1];
            //matrix[x][y] <= matrix[i][j]可以避免使用visited[row][col]数组
            if(x < 0 || x >= row || y < 0 || y >= col || matrix[x][y] <= matrix[i][j]) continue;
            int len = 1 + dfs(matrix, x, y, row, col, cache);
            max = Math.max(len, max);
        }

        cache[i][j] = max;
        return max;
    }
}

[leetcode]longest increasing path in a matrix

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