[LeetCode] Longest Increasing Path in a Matrix 矩阵中的最长递增路径

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.

这道题给我们一个二维数组，让我们求矩阵中最长的递增路径，规定我们只能上下左右行走，不能走斜线或者是超过了边界。那么这道题的解法要用递归和DP来解，用DP的原因是为了提高效率，避免重复运算。我们需要维护一个二维动态数组dp，其中dp[i][j]表示数组中以(i,j)为起点的最长递增路径的长度，初始将dp数组都赋为0，当我们用递归调用时，遇到某个位置(x, y), 如果dp[x][y]不为0的话，我们直接返回dp[x][y]即可，不需要重复计算。我们需要以数组中每个位置都为起点调用递归来做，比较找出最大值。在以一个位置为起点用DFS搜索时，对其四个相邻位置进行判断，如果相邻位置的值大于上一个位置，则对相邻位置继续调用递归，并更新一个最大值，搜素完成后返回即可，参见代码如下：

class Solution {
public:
int longestIncreasingPath(vector<vector<int> >& matrix) {
if (matrix.empty() || matrix[0].empty()) return 0;
int res = 1, m = matrix.size(), n = matrix[0].size();
vector<vector<int> > dp(m, vector<int>(n, 0));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
res = max(res, dfs(matrix, dp, i, j));
}
}
return res;
}
int dfs(vector<vector<int> > &matrix, vector<vector<int> > &dp, int i, int j) {
if (dp[i][j]) return dp[i][j];
vector<vector<int> > dirs = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int mx = 1, m = matrix.size(), n = matrix[0].size();
for (auto a : dirs) {
int x = i + a[0], y = j + a[1];
if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j]) continue;
int len = 1 + dfs(matrix, dp, x, y);
mx = max(mx, len);
}
dp[i][j] = mx;
return mx;
}
};

参考资料：

https://leetcode.com/discuss/81389/15ms-concise-java-solution

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