Minimum Swaps To Make Sequences Increasing：交换元素变为升序

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.

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思路：因为第i个位置是否交换取决于第i-1个位置是否交换， 所以考虑动态规划。交不交换有点类似于背包问题的装与不装。

1. 我们假定n1是 n-1处，没有交换就有序的花费,s1是n-1处交换之后的花费，我们现在要依靠 n1,s1来推测 n2, s2的花费。 
   假设，a1 = A[i-1], b1 = B[i-1] a2 = A[i], b2 = B[i]. 
   如果 a1 < a2 && b1 < b2 则，n2 = min(n1,n2),s2 = min(s2, s1 + 1),这种情况表示要么i-1、i都不交换，要么都要交换，否则不满足升序。
2. a1 < b2 && b1 < a2 n2 = min(n2,s1),s2 = min(s2, n1 + 1), 这种情况表示交换且只能交换i-1或者i。n2表示如果不交换i则只能交换i-1；s2表示不交换i-1，则只能只交换i。

public static int minSwap(int[] A, int[] B) {
    int n1 = 0; //  长度为1时不交换也为升序
    int s1 = 1; // 长度为1时，交换1词后仍为升序

    for (int i = 1; i < A.length; i++) {
        // 以下是推测 n2 s2的过程
        int n2 = Integer.MAX_VALUE;
        int s2 = Integer.MAX_VALUE;
        // 
        if (A[i - 1] < A[i] && B[i - 1] < B[i]) {
            n2 = Math.min(n2, n1);
            s2 = Math.min(s2, s1 + 1);
        }
        if (A[i - 1] < B[i] && B[i - 1] < A[i]) {
            n2 = Math.min(n2, s1);
            s2 = Math.min(s2, n1 + 1);
        }

        n1 = n2;
        s1 = s2;
    }
    return Math.min(n1, s1);
}