[LintCode] Longest Increasing Continuous subsequence II

最新推荐文章于 2023-09-02 10:30:00 发布

转载最新推荐文章于 2023-09-02 10:30:00 发布 · 135 阅读

本文探讨了在给定的二维整数矩阵中寻找最长递增连续子序列的问题，提供了详细的算法实现，包括动态规划方法和深度优先搜索策略，旨在帮助读者理解并解决此类问题。

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Longest Increasing Continuous subsequence II

Give you an integer matrix (with row size n, column size m)，find the longest increasing continuous subsequence in this matrix. (The definition of the longest increasing continuous subsequence here can start at any row or column and go up/down/right/left any direction).

Example

Given a matrix:

[
  [1 ,2 ,3 ,4 ,5],
  [16,17,24,23,6],
  [15,18,25,22,7],
  [14,19,20,21,8],
  [13,12,11,10,9]
]

return 25

Challenge

O(nm) time and memory.

备忘录，dp[i][j] = max{dp[i-1][j], dp[i][j-1], dp[i+1][j], dp[i][j+1]} + 1。

 1 class Solution {
 2 public:
 3     /**
 4      * @param A an integer matrix
 5      * @return  an integer
 6      */
 7     bool isValid(vector<vector<int>> &A, int x, int y) {
 8         return x >= 0 && x < A.size() && y >= 0 && y < A[0].size();
 9     }
10     int dfs(vector<vector<int>> &A, vector<vector<int>> &dp, int x, int y) {
11         if (dp[x][y] != -1) return dp[x][y];
12         const int dx[4] = {0, 1, 0, -1};
13         const int dy[4] = {1, 0, -1, 0};
14         int tmp = 1;
15         for (int i = 0; i < 4; ++i) {
16             int xx = x + dx[i], yy = y + dy[i];
17             if (isValid(A, xx, yy) && A[x][y] > A[xx][yy]) tmp = max(tmp, dfs(A, dp, xx, yy) + 1);
18         }
19         dp[x][y] = tmp;
20         return dp[x][y];
21     }
22     int longestIncreasingContinuousSubsequenceII(vector<vector<int>>& A) {
23         // Write your code here
24         if (A.empty() || A[0].empty()) return 0;
25         int res = 0;
26         vector<vector<int>> dp(A.size(), vector<int>(A[0].size(), -1));
27         for (int i = 0; i < A.size(); ++i) {
28             for (int j = 0; j < A[0].size(); ++j) {
29                 res = max(res, dfs(A, dp, i, j));
30             }
31         }
32         return res;
33     }
34 };