zoukankan      html  css  js  c++  java
  • 765. Couples Holding Hands

    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.

    Approach #1: C++.

    class Solution {
    public:
        int minSwapsCouples(vector<int>& row) {
            int res = 0, N = row.size();
            
            vector<int> ptn(N, 0);
            vector<int> pos(N, 0);
            
            for (int i = 0; i < N; ++i) {
                ptn[i] = (i % 2 == 0 ? i+1 : i-1);
                pos[row[i]] = i;
            }
            
            for (int i = 0; i < N; ++i) {
                for (int j = ptn[pos[ptn[row[i]]]]; i != j; j = ptn[pos[ptn[row[i]]]]) {
                    swap(row[i], row[j]);
                    swap(pos[row[i]], pos[row[j]]);
                    res++;
                }
            }
            
            return res;
        }
    };
    

      

    Analysis:

    https://leetcode.com/problems/couples-holding-hands/discuss/113362/JavaC%2B%2B-O(N)-solution-using-cyclic-swapping

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    SPOJ 10628 求树上的某条路径上第k小的点
    zoj 2112 动态区间求第k大
    SPOJ QTREE 树链剖分
    FZU 2082 过路费
    bzoj 1036 Tree Count
    POJ 3237
    C
    G
    E
    B. Split a Number(字符串加法)
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10305184.html
Copyright © 2011-2022 走看看