leetcode801. Minimum Swaps To Make Sequences Increasing(python)

leetcode801. Minimum Swaps To Make Sequences Increasing(python)

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原题地址：https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/

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: 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].

python代码

class Solution:
    def minSwap(self, A, B):
        """ :type A: List[int] :type B: List[int] :rtype: int """
        dp = [[len(A),len(A)] for i in range(len(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]:
                dp[i][0] = dp[i - 1][0]
                dp[i][1] = dp[i - 1][1] + 1

            if A[i] > B[i - 1] and B[i] > A[i - 1]:
                dp[i][0] = min(dp[i][0], dp[i - 1][1])
                dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1)

        return min(dp[len(A) - 1][0], dp[len(A) - 1][1])

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