LeetCode Top Interview Questions 277. Find the Celebrity (Java版; Medium)

红太狼 2023-07-03 12:05 100阅读 0赞

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## LeetCode Top Interview Questions 277. Find the Celebrity (Java版; Medium) ##

#### 题目描述 ####

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). 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.

![format_png][]

Input: graph = [
      [1,1,0],
      [0,1,0],
      [1,1,1]
    ]
    Output: 1
    Explanation: There are three persons labeled with 0, 1 and 2. graph[i][j] = 1 means person i knows person j,
    otherwise graph[i][j] = 0 means person i does not know person j. The celebrity is the person labeled as 1
    because both 0 and 2 know him but 1 does not know anybody.

![format_png 1][]

Input: graph = [
      [1,0,1],
      [1,1,0],
      [0,1,1]
    ]
    Output: -1
    Explanation: There is no celebrity.
     
    
    Note:
    
    The directed graph is represented as an adjacency matrix, which is an n x n matrix where a[i][j] = 1 means
    person i knows person j while a[i][j] = 0 means the contrary.
    Remember that you won't have direct access to the adjacency matrix.

#### 第一次做; 巧妙的方法; 核心: 找到最可能的候选人; 时间复杂度O(N), 空间复杂度O(1); 这种方法体验一下就可以, 还是要掌握最直观的暴力方法 ####

一种巧妙的办法, 第一步:找到最可能的候选人 第二步:判断该候选人是否是名人
    该方法的核心: 找到最可能的候选人; 不知道数学证明...
    */
    public class Solution extends Relation { 
        public int findCelebrity(int n) { 
            //第一步: 找到最可能的候选人
            int candidate = 0;
            for(int i=1; i<n; i++){ 
                if(knows(candidate, i))
                    candidate = i;
            }
            //第二步: 判断该候选人是否是名人
            for(int i=0; i<n; i++){ 
                if(i==candidate)
                    continue;
                if(knows(i, candidate) && !(knows(candidate, i)))
                    continue;
                return -1;
            }
            return candidate;
        }
    }

#### 第一次做; 暴力的方法, 挨个判断每个人是否是名人; 时间复杂度O(N^2), 空间复杂度O(1) ####

/* The knows API is defined in the parent class Relation. boolean knows(int a, int b); */
    /* 暴力的方法, 挨个判断当前这个人是否是名人, 这也是最直观的方法 细节: 自己认识自己 */
    public class Solution extends Relation { 
        public int findCelebrity(int n) { 
            for(int i=0; i<n; i++){ 
                //判断i是不是名人, 看看是不是满足名人的定义:1)剩下n-1个人认识i, 2)i不认识他们中的任何一个
                int j=0;
                for(; j<n; j++){ 
                    if(i==j)
                        continue;
                    //j认识i 同时 i不认识j
                    if(knows(j, i) && !knows(i, j))
                        continue;
                    break;
                }
                //如果j==n, 说明1)剩下n-1个人认识i, 2)i不认识他们中的任何一个; i是名人
                if(j==n)
                    return i;
            }
            return -1;
        }
    }

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