LeetCode[329] Longest Increasing Path in a Matrix

cppowboy 发布于2019-08-14 15:01 / 729人阅读

摘要：复杂度思路为了避免搜索已经搜索的点。所以考虑用一个数组，记录到每一个点能产生的最长序列的长度。考虑用进行搜索，对于每一个点来说，考虑先找到最小的那个点针对每一条路径的。然后对于每一点再递增回去，依次累积找到增加的值。

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.

DP + DFS

复杂度
O(MN),O(N)

思路
为了避免搜索已经搜索的点。所以考虑用一个dp数组，记录到每一个点能产生的最长序列的长度。考虑用dfs进行搜索，对于每一个点来说，考虑先找到最小的那个点针对每一条路径的。然后对于每一点再递增回去，依次累积找到增加的值。

代码

public int longestIncreasingPath(int[][] matrix) {
    if(matrix.length == 0) return 0; 
    int row = matrix.length, col = matrix[0].length;
    int[][] dp = new int[row][col];
    //dp = {0};
    int val = 0;
    for(int i = 0; i < row; i ++) {
        for(int j = 0; j < col; j ++) {
            if(helper(matrix, i, j, dp) > val) {
                val = helper(matrix, i, j, dp);
            }
        }
    }
    return val;
}

public int helper(int[][] matrix, int i, int j, int[][] dp) {
    //this is the point;
    if(dp[i][j] != 0) return dp[i][j];
    int val = 0;
    if(i + 1 < matrix.length && matrix[i + 1][j] < matrix[i][j]) {
        val = Math.max(val, helper(matrix, i + 1, j, dp));
    }
    if(i - 1 >= 0 && matrix[i - 1][j] < matrix[i][j]) {
        val = Math.max(val, helper(matrix, i - 1, j, dp));
    }
    if(j + 1 < matrix[0].length && matrix[i][j + 1] < matrix[i][j]) {
        val = Math.max(val, helper(matrix, i, j + 1, dp));
    }
    if(j - 1 >= 0 && matrix[i][j - 1] < matrix[i][j]) {
        val = Math.max(val, helper(matrix, i, j - 1, dp));
    }
    dp[i][j] = val + 1;
    return dp[i][j];
}

GPU云服务器云服务器 Increasing Longest Matrix modelview matrix

文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。

转载请注明本文地址：https://www.ucloud.cn/yun/65237.html

leetcode 329. Longest Increasing Path in a Matrix

摘要：题目要求思路和代码这里采用广度优先算法加上缓存的方式来实现。我们可以看到，以一个节点作为开始构成的最长路径长度是确定的。因此我们可以充分利用之前得到的结论来减少重复遍历的次数。题目要求 Given an integer matrix, find the length of the longest increasing path. From each cell, you can ei...

heartFollower 2019-08-16 11:09 评论0 收藏0
[LeetCode] 329. Longest Increasing Path in a Matri

Problem 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 ou...

hss01248 2019-08-16 13:25 评论0 收藏0
[LeetCode] 329. Longest Increasing Path in a Matri

Problem 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 ou...

antz 2019-08-06 12:26 评论0 收藏0
LeetCode 329. Longest Increasing Path in a Matrix

摘要：思路这道题主要使用记忆化递归和深度优先遍历。我们以给定的矩阵的每一个位置为起点，进行深度优先遍历。我们存储每个位置深度优先遍历的结果，当下一次走到这个位置的时候，我们直接返回当前位置记录的值，这样可以减少遍历的次数，加快执行速度。 Description Given an integer matrix, find the length of the longest increasing...

isLishude 2019-07-31 10:08 评论0 收藏0
329. Longest Increasing Path in a Matrix

摘要：题目解答最重要的是用保存每个扫过结点的最大路径。我开始做的时候，用全局变量记录的没有返回值，这样很容易出错，因为任何一个用到的环节都有可能改变值，所以还是在函数中定义，把当前的直接返回计算不容易出错。题目：Given an integer matrix, find the length of the longest increasing path. From each cell, y...

hqman 2019-08-14 14:17 评论0 收藏0