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

2021年09月15日 阅读数：2

We have two integer sequences `A` and `B` of the same non-zero length.html

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.java

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]`.)python

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.数组

```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:post

• `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]`.

Java:code

```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:htm

```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++:blog

```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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