链接:https://leetcode.com/problems/couples-holding-hands/description/
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:
len(row)
is even and in the range of[4, 60]
.row
is guaranteed to be a permutation of0...len(row)-1
.
思路:
对于row[0]来说,如果row[1] 和 row[0]是夫妻,那么他们两个就都不动就好。如果两个都移走,那么不是最优。
考虑row[0]和row[1]不是夫妻的情况
操作1: 要么把row[0]和 row[0]的夫妻边上的交换
操作2: 要么把row[1]和row[0]的夫妻交换
操作3:要么row[0]和row[1]都移走。
显然操作3要不得。
对于0和1间隔奇数的情况:
0abc1d
变成 dabc10 或 01bcda
对于0和1间隔偶数的情况
0abcd1
变成 01bcda 或者 dabc01
class Solution { public: int minSwapsCouples(vector<int>& row) { int n = row.size(); for (int i = 0; i < n; i++) { row[i] -= row[i] % 2; } int ans = 0; for (int i = 0; i < n; i += 2) { if (row[i] == row[i + 1]) { continue; } for (int j = i + 2; j < n; j++) { if (row[i] == row[j]) { ans++; swap(row[i + 1], row[j]); break; } } } return ans; } };