801. Minimum Swaps To Make Sequences Increasing

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]

.

题意：

给出两个相同长度的数组，求至少交换几次，使得两个数组都是严格递增的。交换必须是相同的index处交换，即A[i]只能与B[i]交换。

思路：

答案都看半天，手动再见

，真难啊。。。还是找状态转移方程，设置数组dp[i][0]和dp[i][1]，dp[i][0]表示第i处不交换需要交换的总次数，dp[i][1]表示第i次交换所需要交换的总次数。如果A[i-1]<A[i]&&B[i-1]<B[i]，那么第i处不交换的话，dp[i][[0=min(dp[i][0],dp[i-1][0]),交换的话，dp[i][1]=min(dp[i-1][1]+1,dp[i][1])，因为交换第i处，那么第i-1处也可能也要交换。
对于A[i-1]<B[i]&&B[i-1]<A[i]，也可以照此分析。

代码:

class Solution {
public int minSwap(int[] A, int[] B) {
int [][]dp=new int[A.length][2];
for(int i=0;i<A.length;i++)
{
dp[i][0]=Integer.MAX_VALUE;
dp[i][1]=Integer.MAX_VALUE;
}
dp[0][0]=0;
dp[0][1]=1;
for(int i=1;i<A.length;i++)
{
if(A[i-1]<A[i]&&B[i-1]<B[i])
dp[i][0]=Math.min(dp[i][0],dp[i-1][0]);
if(A[i-1]<B[i]&&B[i-1]<A[i])
dp[i][0]=Math.min(dp[i][0],dp[i-1][1]);
if(A[i-1]<A[i]&&B[i-1]<B[i])
dp[i][1]=Math.min(dp[i-1][1]+1,dp[i][1]);
if(A[i-1]<B[i]&&B[i-1]<A[i])
dp[i][1]=Math.min(dp[i][1],dp[i-1][0]+1);
}
return Math.min(dp[A.length-1][0],dp[A.length-1][1]);
}
}