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

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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