Description

We have two integer sequences A and B of the same non-zero length.

We are allowed to swap elements A[i] and B[i]. Note that both elements are in the same index position in their respective sequences.

At the end of some number of swaps, A and B are both strictly increasing. (A sequence is strictly increasing if and only if A[0] < A[1] < A[2] < … < A[A.length - 1].)

Given A and B, return the minimum number of swaps to make both sequences strictly increasing. It is guaranteed that the given input always makes it possible.

Example:

Input: A = [1,3,5,4], B = [1,2,3,7]
Output: 1
Explanation: 
Swap A[3] and B[3].  Then the sequences are:
A = [1, 3, 5, 7] and B = [1, 2, 3, 4]
which are both strictly increasing.

Note:

  • A, B are arrays with the same length, and that length will be in the range [1, 1000].
  • A[i], B[i] are integer values in the range [0, 2000].

分析

题目的意思是：求使两个序列递增的最小交换次数。

用二维DP来计算：

  • dp[i][0]表示不交换i，使得[0, i]严格递增的最小swap数
  • dp[i][1]表示交换i，使得[0, i]严格递增的最小swap数
  • 再看状态转移方程，在每一步判断，我们要不要交换，A[i-1]<B[i]&&B[i-1]<A[i]时这个是必须要交换的；而A[i-1]<A[i]&&B[i-1]<B[i]也可以交换，也可以不交换。

代码

class Solution {
public:
    int minSwap(vector<int>& A, vector<int>& B) {
        int n=A.size();
        vector<vector<int>> dp(n,vector<int>(2,INT_MAX));
        dp[0][0]=0;
        dp[0][1]=1;
        for(int i=1;i<n;i++){
            if(A[i-1]<A[i]&&B[i-1]<B[i]){
                dp[i][0]=min(dp[i][0],dp[i-1][0]);
                dp[i][1]=min(dp[i-1][1]+1,dp[i][1]);
            }
            if(A[i-1]<B[i]&&B[i-1]<A[i]){
                dp[i][0]=min(dp[i-1][1],dp[i][0]);
                dp[i][1]=min(dp[i-1][0]+1,dp[i][1]);
            }
        }
        return min(dp[n-1][0],dp[n-1][1]);
    }
};

参考文献

​​801. Minimum Swaps To Make Sequences Increasing​​