329. Longest Increasing Path in a Matrix 矩阵中的最长递增路径
作者: 负雪明烛 id: fuxuemingzhu 个人博客: https://fuxuemingzhu.cn/
题目地址: https://leetcode.com/problems/longest-increasing-path-in-a-matrix/description/
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: 4Explanation: The longest increasing path is [1, 2, 6, 9].Example 2:
Input: nums =[ [3,4,5], [3,2,6], [2,2,1]]Output: 4Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.求二维矩阵中最长的递增路径。
和417. Pacific Atlantic Water Flow非常类似,直接DFS求解。一般来说DFS需要有固定的起点,但是对于这个题,二维矩阵中的每个位置都算作起点。
把每个位置都当做起点,然后去做个dfs,看最长路径是多少。然后再找出全局的最长路径。使用cache保存已经访问过的位置,这样能节省了很多搜索的过程,然后有个continue是为了剪枝。因为这个做法比较暴力,就没有什么好讲的了。
最坏情况下的时间复杂度是O((MN)^2),空间复杂度是O(MN)。
class Solution(object): def longestIncreasingPath(self, matrix): """ :type matrix: List[List[int]] :rtype: int """ if not matrix or not matrix[0]: return 0 m, n = len(matrix), len(matrix[0]) res = 0 cache = [[-1] * n for _ in range(m)] for i in range(m): for j in range(n): path = self.dfs(matrix, cache, m, n, i, j) res = max(res, path) return res
def dfs(self, matrix, cache, m, n, i, j): if cache[i][j] != -1: return cache[i][j] directions = [(-1, 0), (1, 0), (0, 1), (0, -1)] res = 1 for dire in directions: x, y = i + dire[0], j + dire[1] if x < 0 or x >= m or y < 0 or y >= n or matrix[x][y] <= matrix[i][j]: continue path = 1 + self.dfs(matrix, cache, m, n, x, y) res = max(path, res) cache[i][j] = res return cache[i][j]参考资料:
2018 年 10 月 1 日 —— 欢度国庆!

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