[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:

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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:

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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++:

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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++:

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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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