【CODE】Couples Holding Hands

765. Couples Holding Hands

Hard

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N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats.

The people and seats are represented by an integer from 0 to 2N-1, the couples are numbered in order, the first couple being (0, 1), the second couple being (2, 3), and so on with the last couple being (2N-2, 2N-1).

The couples' initial seating is given by row[i] being the value of the person who is initially sitting in the i-th seat.

Example 1:

Input: row = [0, 2, 1, 3]
Output: 1
Explanation: We only need to swap the second (row[1]) and third (row[2]) person.

Example 2:

Input: row = [3, 2, 0, 1]
Output: 0
Explanation: All couples are already seated side by side.

Note:

1. len(row) is even and in the range of [4, 60].
2. row is guaranteed to be a permutation of 0...len(row)-1.

• 同样运用交换的手法：(1,0) or (0,1) 是一对，即每对奇数是较大的，遍历，根据数字是奇数还是偶数，判断他的下一个数字是否与他是一对，遍历时，i 的步长是 2。
• Runtime: 0 ms, faster than 100.00% of C++ online submissions for Couples Holding Hands.
• Memory Usage: 8.5 MB, less than 61.54% of C++ online submissions for Couples Holding Hands.
• Next challenges:
• K-Similar Strings

class Solution {
public:
    int minSwapsCouples(vector<int>& row) {
        int n=row.size();
        int i;
        int res=0;
        for(i=0;i<n-1;i+=2){
            if(row[i]%2==1 && row[i+1]==row[i]-1 || row[i]%2==0 && row[i+1]==row[i]+1) continue;
            else{
                int tar=0;
                if(row[i]%2==1) tar=row[i]-1;
                else tar=row[i]+1;
                for(int j=i+2;j<n;j++){
                    if(row[j]==tar){
                        swap(row[j],row[i+1]);
                        res++;
                        break;
                    }
                }
            }
        }
        return res;
    }
};

• 另外看一下：https://www.cnblogs.com/grandyang/p/8716597.html