leetcode笔记:Longest Increasing Path in a Matrix
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一. 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.
二. 题目分析
题目的大意是,给定一个整数矩阵,计算其要求元素排列是递增的,球最长递增路径的长度。
从任意一个矩阵位置出发,可向上下左右四个方向移动。不可以沿着对角线移动,也不能离开边界。(环绕也是不允许的)。题目还给出了两个测试用例。
解题思路是,深度优先搜索,但如果处理不好也可能超时,你需要加入记忆化搜索,具体方法如下:
从起点开始遍历矩阵,递归寻找其最长递增路径。定义辅助数组record,用于记录已经搜索过的矩阵元素的数据,如record[x][y]
存储了从坐标(x, y)
出发的最长递增路径长度。
之后,进行深度优先搜索。逐一检查某个元素的四个方向,并继续从所有可能出现最长路径的方向上进行搜索。 当遇到record[x][y]
已算出的情况,直接返回record[x][y]
,减少运算量。
三. 示例代码
class Solution {private: int dfs(vector<vector<int>>& matrix, vector<vector<int>>& record, int x, int y, int lastVal) { if (x < 0 || y < 0 || x >= matrix.size() || y >= matrix[0].size()) return 0; if (matrix[x][y] > lastVal) { if (record[x][y] != 0) return record[x][y]; // 路线已算出,直接返回结果 int left = dfs(matrix, record, x + 1, y, matrix[x][y]) + 1; int right = dfs(matrix, record, x - 1, y, matrix[x][y]) + 1; int top = dfs(matrix, record, x, y + 1, matrix[x][y]) + 1; int bottom = dfs(matrix, record, x, y - 1, matrix[x][y]) + 1; record[x][y] = max(left, max(right, max(top, bottom))); return record[x][y]; } return 0; }public: int longestIncreasingPath(vector<vector<int>>& matrix) { if (matrix.size() == 0) return 0; vector<int> temp(matrix[0].size(), 0); vector<vector<int>> record(matrix.size(), temp); int longest = 0; for (int i = 0; i < matrix.size(); ++i) for (int j = 0; j < matrix[0].size(); ++j) longest = max(longest, dfs(matrix, record, i, j, -1)); return longest; }};
四. 小结
该题是DFS + 记忆化搜索的经典题目。
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