[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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