Longest Continuous Increasing Subsequence II

Description

Given an integer matrix. Find the longest increasing continuous subsequence in this matrix and return the length of it.

The longest increasing continuous subsequence here can start at any position and go up/down/left/right.

Example

Example 1:

Input:
    [
      [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]
    ]
Output: 25
Explanation: 1 -> 2 -> 3 -> 4 -> 5 -> ... -> 25 (Spiral from outside to inside.)

Example 2:

Input:
    [
      [1, 2],
      [5, 3]
    ]
Output: 4
Explanation: 1 -> 2 -> 3 -> 5

Challenge

Assume that it is a N x M matrix. Solve this problem in O(NM) time and memory.

思路：

动态规划, 设定状态 f[i][j] 表示矩阵中坐标 (i, j) 的点开始的最长上升子序列

状态转移方程:

int dx[4] = {0, 1, -1, 0};
int dy[4] = {1, 0, 0, -1};

f[i][j] = max{ f[i + dx[k]][j + dy[k]] + 1 }

k = 0, 1, 2, 3, matrix[i + dx[k]][j + dy[k]] > matrix[i][j]

这道题目可以向四个方向走, 所以推荐使用记忆化搜索(递归)的写法.

(当然, 也可以反过来设定: f[i][j] 表示走到 (i, j) 的最长上升子序列, 相应的状态转移方程做一点点改变即可)

public class Solution {
    /**
     * @param matrix: A 2D-array of integers
     * @return: an integer
     */
     int[][] dp;
    int n, m;

    public int longestContinuousIncreasingSubsequence2(int[][] A) {
        if (A.length == 0) {
            return 0;
        }

        n = A.length;
        m = A[0].length;
        int ans = 0;
        dp = new int[n][m]; // dp[i][j] means the longest continuous increasing path from (i,j)
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                dp[i][j] = -1; // dp[i][j] has not been calculated yet
            }
        }

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                search(i, j, A);
                ans = Math.max(ans, dp[i][j]);
            }
        }

        return ans;
    }

    int[] dx = { 1, -1, 0, 0 };
    int[] dy = { 0, 0, 1, -1 };

    void search(int x, int y, int[][] A) {
        if (dp[x][y] != -1) { // if dp[i][j] has been calculated, return directly
            return;
        }

        int nx, ny;
        dp[x][y] = 1;
        for (int i = 0; i < 4; ++i) {
            nx = x + dx[i];
            ny = y + dy[i];
            if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
                if (A[nx][ny] > A[x][y]) {
                    search(nx, ny, A); // dp[nx][ny] must be calcuted
                    dp[x][y] = Math.max(dp[x][y], dp[nx][ny] + 1);
                }
            }
        }
    }
}