[Leetcode] 765. Couples Holding Hands 解题报告

题目：

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.

思路：

采用贪心的思路：为了便于快速查找某个id的人所处的位置，我们定义一个哈希表，表示从某人的id到其在row中的位置。然后就开始扫描row中的偶数位。根据在该偶数位上的人的id，可以推出其另一半所处的位置。如果他的另一半不在他接下来的位置，那么我们就交换下一位置和它的另一半的位置（并且记得更新哈希表）。这样最后返回总共的交换次数即可。

在数学上可以漂亮的证明这种贪心策略是正确的，具体可以参见Cyclic permutation。我们实现的算法的时间复杂度为O(n)，空间复杂度也为O(n)。

代码：

class Solution {
public:
    int minSwapsCouples(vector<int>& row) {
        unordered_map<int, int> hash;   // map from person id to position
        for (int i = 0; i < row.size(); ++i) {
            hash[row[i]] = i;
        }
        int ret = 0;
        for (int i = 0; i < row.size(); i += 2) {
            int p1 = row[i];
            int p2 = p1 % 2 == 0 ? p1 + 1 : p1 - 1; // its couple's id
            if (row[i + 1] != p2) {
                int p2_pos = hash[p2];
                hash[row[i + 1]] = p2_pos;
                hash[p2] = i + 1;
                swap(row[i + 1], row[p2_pos]);
                ++ret;
            }
        }
        return ret;
    }
};