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.
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
思路:因为第i个位置是否交换取决于第i-1个位置是否交换, 所以考虑动态规划。交不交换有点类似于背包问题的装与不装。
- 我们假定
n1
是n-1
处,没有交换就有序的花费,s1
是n-1
处交换之后的花费,我们现在要依靠n1
,s1
来推测n2
,s2
的花费。
假设,a1 = A[i-1], b1 = B[i-1]
a2 = A[i], b2 = B[i]
.
如果a1 < a2
&&b1 < b2
则,n2 = min(n1,n2),s2 = min(s2, s1 + 1)
,这种情况表示要么i-1、i都不交换,要么都要交换,否则不满足升序。 a1 < b2 && b1 < a2
n2 = min(n2,s1),s2 = min(s2, n1 + 1)
, 这种情况表示交换且只能交换i-1或者i。n2表示如果不交换i则只能交换i-1;s2表示不交换i-1,则只能只交换i。
public static int minSwap(int[] A, int[] B) {
int n1 = 0; // 长度为1时不交换也为升序
int s1 = 1; // 长度为1时,交换1词后仍为升序
for (int i = 1; i < A.length; i++) {
// 以下是推测 n2 s2的过程
int n2 = Integer.MAX_VALUE;
int s2 = Integer.MAX_VALUE;
//
if (A[i - 1] < A[i] && B[i - 1] < B[i]) {
n2 = Math.min(n2, n1);
s2 = Math.min(s2, s1 + 1);
}
if (A[i - 1] < B[i] && B[i - 1] < A[i]) {
n2 = Math.min(n2, s1);
s2 = Math.min(s2, n1 + 1);
}
n1 = n2;
s1 = s2;
}
return Math.min(n1, s1);
}