801. Minimum Swaps To Make Sequences Increasing

问题描述：

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].

 

解题思路：

参考了https://www.jianshu.com/p/6021550d227b的解释

可以用动态规划来解答。

类似于背包问题。背包问题为取或不取

而这个问题是，换或不换。

swap[i] 表示换第i个最小的次数

nswap[i]表示不换第i个最小的交换次数。

 

当A[i-1] < A[i] 以及 B[i-1] < B[i]时，此时若前i-1为严格递增，则此时一定为严格递增。

要么i-1和i位置都不换，要么都换。

所以swap[i] = swap[i-1]+1 ; nswap[i] = nswap[i-1];

考虑只换i的情况，此时应满足：A[i-1] < B[i] && B[i-1] < A[i]

此时：swap[i] = min(swap[i], nswap[i-1] + 1);  nswap[i] = min(nswap[i], swap[i-1]);

 

代码：

class Solution {
public:
    int minSwap(vector<int>& A, vector<int>& B) {
        int N = A.size();
        if(N == 0) return 0;
        int nswap[1000] = {0};
        int swap[1000] = {1};
        for(int i = 1; i < N; i++){
            nswap[i] = swap[i] = N;
            if(A[i-1] < A[i] && B[i-1] < B[i]){
                nswap[i] = nswap[i-1];
                swap[i] = swap[i-1]+1;
            }
            if(A[i-1] < B[i] && B[i-1] < A[i]){
                nswap[i] = min(nswap[i], swap[i-1]);
                swap[i] = min(swap[i], nswap[i-1]+1);
            }
        }
        return min(swap[N-1], nswap[N-1]);
    }
};