【leetcode】801. Minimum Swaps To Make Sequences Increasing

题目如下：

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：每个下标对应的元素只有交换和不交换两种选择，记dp[i][0]为在[0~i]这个区间内，在第i个元素不交换时使得[0~i]区间子数组严格递增时总的交换次数，而dp[i][0]为在[0~i]这个区间内，在第i个元素交换时使得[0~i]区间子数组严格递增时总的交换次数。要使得数组严格递增，第i个元素是否需要交换取决于与(i-1)元素的值的大小情况，总得来说分为可能性如下，

1.  A[i] > A[i - 1] and B[i] > B[i - 1]  and A[i] > B[i - 1] and B[i] > A[i - 1] ，这种情况下，第i个元素可以交换或者不交换，并且和i-1是否交换没有任何关系，那么可以得出：  在第i个元素不交换的情况下，dp[i][0] 应该等于第i-1个元素交换与不交换两种情况下的较小值，有 dp[i][0] = min(dp[i][0], dp[i - 1][0], dp[i - 1][1])  ，如果第i个元素非要任性的交换，那么结果就是第i-1个元素交换与不交换两种情况下的较小值加上1，有dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1, dp[i - 1][1] + 1) 。

2. A[i] > A[i - 1] and B[i] > B[i - 1] ，这种情况是i和i-1之间要么都交换，要么都不交换。有 dp[i][0] = min(dp[i][0], dp[i - 1][0]) ，dp[i][1] = min(dp[i][1], dp[i - 1][1] + 1)

3. A[i] > B[i - 1] and B[i] > A[i - 1] and (A[i] <= A[i - 1] or B[i] <= B[i - 1])，这种情况是要么i交换，要么i-1交换。有 dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1)，dp[i][0] = min(dp[i][0], dp[i - 1][1])

4.其他情况则表示无论i交换或者不交换都无法保证严格递增。

代码如下：

class Solution(object):
    def minSwap(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: int
        """
        dp = [[float('inf')] * 2 for _ in A]
        dp[0][0] = 0
        dp[0][1] = 1
        for i in range(1, len(A)):
            if (A[i] > A[i - 1] and B[i] > B[i - 1]) and (A[i] > B[i - 1] and B[i] > A[i - 1]):
                dp[i][0] = min(dp[i][0], dp[i - 1][0], dp[i - 1][1])
                dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1, dp[i - 1][1] + 1)
            elif A[i] > A[i - 1] and B[i] > B[i - 1]:
                dp[i][0] = min(dp[i][0], dp[i - 1][0])
                dp[i][1] = min(dp[i][1], dp[i - 1][1] + 1)
            elif A[i] > B[i - 1] and B[i] > A[i - 1] and (A[i] <= A[i - 1] or B[i] <= B[i - 1]):
                dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1)
                dp[i][0] = min(dp[i][0], dp[i - 1][1])

        #print dp
        return min(dp[-1]) if min(dp[-1]) != float('inf') else -1

原文地址：https://www.cnblogs.com/seyjs/p/11737123.html