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].

给两个长度相等的数组A和B，可在任意位置i交换A[i]和B[i]的值，使得数组A和B变成严格递增的数组，求最少需要交换的次数。

解法：dp

Java:

class Solution {

    public int minSwap(int[] A, int[] B) {

        int swapRecord = 1, fixRecord = 0;

        for (int i = 1; i < A.length; i++) {

            if (A[i - 1] >= B[i] || B[i - 1] >= A[i]) {

		// In this case, the ith manipulation should be same as the i-1th manipulation

                // fixRecord = fixRecord;

                swapRecord++;

            } else if (A[i - 1] >= A[i] || B[i - 1] >= B[i]) {

		// In this case, the ith manipulation should be the opposite of the i-1th manipulation

                int temp = swapRecord;

                swapRecord = fixRecord + 1;

                fixRecord = temp;

            } else {

                // Either swap or fix is OK. Let's keep the minimum one

                int min = Math.min(swapRecord, fixRecord);

                swapRecord = min + 1;

                fixRecord = min;

            }

        }

        return Math.min(swapRecord, fixRecord);

    }

}

Python:

class Solution(object):

    def minSwap(self, A, B):

        """

        :type A: List[int]

        :type B: List[int]

        :rtype: int

        """

        dp_no_swap, dp_swap = [0]*2, [1]*2

        for i in xrange(1, len(A)):

            dp_no_swap[i%2], dp_swap[i%2] = float("inf"), float("inf")

            if A[i-1] < A[i] and B[i-1] < B[i]:

                dp_no_swap[i%2] = min(dp_no_swap[i%2], dp_no_swap[(i-1)%2])

                dp_swap[i%2] = min(dp_swap[i%2], dp_swap[(i-1)%2]+1)

            if A[i-1] < B[i] and B[i-1] < A[i]:

                dp_no_swap[i%2] = min(dp_no_swap[i%2], dp_swap[(i-1)%2])

                dp_swap[i%2] = min(dp_swap[i%2], dp_no_swap[(i-1)%2]+1)

        return min(dp_no_swap[(len(A)-1)%2], dp_swap[(len(A)-1)%2])

C++:

class Solution {

public:

    int minSwap(vector<int>& A, vector<int>& B) {

        int n = A.size();

        vector<int> swap(n, n), noSwap(n, n);

        swap[0] = 1; noSwap[0] = 0;

        for (int i = 1; i < n; ++i) {

            if (A[i] > A[i - 1] && B[i] > B[i - 1]) {

                swap[i] = swap[i - 1] + 1;

                noSwap[i] = noSwap[i - 1];

            }

            if (A[i] > B[i - 1] && B[i] > A[i - 1]) {

                swap[i] = min(swap[i], noSwap[i - 1] + 1);

                noSwap[i] = min(noSwap[i], swap[i - 1]);

            }

        }

        return min(swap[n - 1], noSwap[n - 1]);

    }

};

C++:

class Solution {

public:

    int minSwap(vector<int>& A, vector<int>& B) {

        int n1 = 0, s1 = 1, n = A.size();

        for (int i = 1; i < n; ++i) {

            int n2 = INT_MAX, s2 = INT_MAX;

            if (A[i - 1] < A[i] && B[i - 1] < B[i]) {

                n2 = min(n2, n1);

                s2 = min(s2, s1 + 1);

            }

            if (A[i - 1] < B[i] && B[i - 1] < A[i]) {

                n2 = min(n2, s1);

                s2 = min(s2, n1 + 1);

            }

            n1 = n2;

            s1 = s2;

        }

        return min(n1, s1);

    }

};

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[LeetCode] 801. Minimum Swaps To Make Sequences Increasing 最少交换使得序列递增

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