[LintCode] Longest Increasing Continuous subsequence II

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。

class Solution {
public:
    /**
     * @param A an integer matrix
     * @return  an integer
     */
    bool isValid(vector<vector<int>> &A, int x, int y) {
        return x >= 0 && x < A.size() && y >= 0 && y < A[0].size();
    }
    int dfs(vector<vector<int>> &A, vector<vector<int>> &dp, int x, int y) {
        if (dp[x][y] != -1) return dp[x][y];
        const int dx[4] = {0, 1, 0, -1};
        const int dy[4] = {1, 0, -1, 0};
        int tmp = 1;
        for (int i = 0; i < 4; ++i) {
            int xx = x + dx[i], yy = y + dy[i];
            if (isValid(A, xx, yy) && A[x][y] > A[xx][yy]) tmp = max(tmp, dfs(A, dp, xx, yy) + 1);
        }
        dp[x][y] = tmp;
        return dp[x][y];
    }
    int longestIncreasingContinuousSubsequenceII(vector<vector<int>>& A) {
        // Write your code here
        if (A.empty() || A[0].empty()) return 0;
        int res = 0;
        vector<vector<int>> dp(A.size(), vector<int>(A[0].size(), -1));
        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < A[0].size(); ++j) {
                res = max(res, dfs(A, dp, i, j));
            }
        }
        return res;
    }
};