题目描述：

LeetCode 801. Minimum Swaps To Make Sequences Increasing

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].

题目大意：

给定数组A与B，通过若干次交换对应下标的元素A[i]与B[i]，使得A与B分别严格递增有序。

求最少的交换次数。

解题思路：

动态规划（Dynamic Programming）

swap和keep分别记录A[i]与B[i]交换或者不交换时的最小代价

分三种情况讨论：

A[i]与B[i]必须交换

A[i]与B[i]可以交换也可以保持原状

A[i]与B[i]必须保持原状

Python代码：

class Solution(object):
    def minSwap(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: int
        """
        swap, keep = 1, 0
        for i in range(1, len(A)):
            if A[i] <= A[i - 1] or B[i] <= B[i - 1]:
                # swap
                nswap = keep + 1
                nkeep = swap
            elif A[i] > B[i - 1] and B[i] > A[i - 1]:
                # swap or keep
                nkeep = min(keep, swap)
                nswap = nkeep + 1
            else:
                # keep
                nkeep = keep
                nswap = swap + 1
            swap, keep = nswap, nkeep
        return min(swap, keep)

 

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