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.

class Solution {
public:
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;
}

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);//record[x][y]存储了从坐标(x, y)出发的最长递增路径长度
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;
}
};