zoukankan      html  css  js  c++  java
  • 277. Find the Celebrity

    https://leetcode.com/problems/find-the-celebrity/#/description

    Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the other n - 1 people know him/her but he/she does not know any of them.

    Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B. You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).

    You are given a helper function bool knows(a, b) which tells you whether A knows B. Implement a function int findCelebrity(n), your function should minimize the number of calls to knows.

    Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return -1.

    Sol 1:

    Perfect problem to train loop structure.

    Brute-force. 

    For each person, check if other people know him or he knows other people.

    We will implement 2 "for" loops. Skip when the outer loop equals to the inner loop because they are the same person.  

    # The knows API is already defined for you.
    # @param a, person a
    # @param b, person b
    # @return a boolean, whether a knows b
    # def knows(a, b):
    
    class Solution(object):
        def findCelebrity(self, n):
            """
            :type n: int
            :rtype: int
            """
            
            # Brute-Force
            # use for - else construct. 
            for i in range(n):
                for j in range(n):
                    if i == j:
                        continue
                    if [knows(i,j), knows(j,i)] != [False, True]:
                        break
                else:
                    return i 
            return -1
            

    Note:

    1 for - else:

    else part is excecuted when for is not ended by break.  

  • 相关阅读:
    js封装一个哈希表
    js封装一个双链表
    js封装一个单链表
    js封装一个栈
    js封装一个优先级队列
    js封装一个队列
    微信小程序开发中自定义自适应头部导航栏
    Git的基本使用
    6位半数字万用表解释
    内存相关概念详解
  • 原文地址:https://www.cnblogs.com/prmlab/p/7120454.html
Copyright © 2011-2022 走看看