原题链接在这里：

题目：

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: 1Explanation: 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].

题解：

Let swap denotes minimum swaps till index i ending with swapping A[i] and B[i].

Let fix denotes minimum swaps till index i ending with NOT swapping A[i] and B[i].

There would be 3 cases:

case1, A[i] <= B[i-1] || B[i] <= A[i-1], i step choice should be same as i-1 step. If i-1 use swap, i need swap. If i-1 use fix, i need fix. Otherwise, i swap would make the result invalid.

case 2, A[i] <= A[i-1] || B[i] <= B[i-1], i step choice should be opposite from i-1 step. If i-1 use swap, i need fix. If i-1 use fix, i need swap. Otherwise, if both swap, the result would be invalid.

case 3, all others, swap or fix are both okay. Then swap takes previous min(swap, fix) plus 1. fix takes previous min(swap, fix) and no change.

Time Complexity: O(A.length).

Space: O(1).

AC Java:

1 class Solution { 2     public int minSwap(int[] A, int[] B) { 3         if(A == null || B == null || A.length < 2){ 4             return 0; 5         } 6          7         int fix = 0; 8         int swap = 1; 9         for(int i = 1; i