[leetcode] 329. Longest Increasing Path in a Matrix 解题报告

题目链接：https://leetcode.com/problems/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.

思路：DFS + 记忆化搜索

使用一个数组来记录已经搜索过的路径。调起来还是花了一些时间。

代码如下：

class Solution {
public:
int DFS(vector<vector<int>>& matrix, int y, int x, int val, vector<vector<int>>& hash)
{
if(y < 0 || y >= matrix.size() || x <0 || x >= matrix[0].size())
return 0;
if(matrix[y][x] > val)
{
if(hash[y][x] != 0) return hash[y][x]; //if this path has been searched
int a = DFS(matrix, y, x+1,matrix[y][x], hash) + 1;
int b = DFS(matrix, y, x-1,matrix[y][x], hash) + 1;
int c = DFS(matrix, y+1, x,matrix[y][x], hash) + 1;
int d = DFS(matrix, y-1, x,matrix[y][x], hash) + 1;
hash[y][x] = max(a, max(b,max(c, d)));
return hash[y][x];
}
return 0;
}
int longestIncreasingPath(vector<vector<int>>& matrix) {
if(matrix.size() == 0) return 0;
int Max = 0;
vector<vector<int>> hash(matrix.size(), vector<int>(matrix[0].size(), 0));
for(int i = 0; i< matrix.size(); i++)
for(int j = 0; j < matrix[0].size(); j++)
Max = max(DFS(matrix, i, j, INT_MIN, hash), Max);
return Max;
}
};