Find the Celebrity

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.

Example 1:

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.

Example 2:

Input: graph = [
  [1,0,1],
  [1,1,0],
  [0,1,1]
]
Output: -1
Explanation: There is no celebrity.

Note:

1. 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.

2. Remember that you won't have direct access to the adjacency matrix.

分析

这里一轮就能出candidate, 然后横纵方向验证。注意row有个例外i!= candidate 一定排除，在python里直接让range(candidate)排除该选项。

/* The knows API is defined in the parent class Relation.
      boolean knows(int a, int b); */

public class Solution extends Relation {
    public int findCelebrity(int n) {
        int x = 0;
        for(int j = 1; j < n; j++){
            if(knows(x,j) ){
                x = j;
            }
        }
        for(int i = 0; i <n; i ++){
            if(knows(x,i) && i!=x || !knows(i,x)) //这里看矩阵，比较row时候x=i是例外，要排除
                return -1;
            }
            
        
        return x;
    }
}

python

# 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
        """
        cand = 0
        for i in range(1,n):
            if knows(cand,i):
                cand = i
        #knows(cand,i)是判断row,要排除i = candidate情况。
        if any(knows(cand,i) for i in xrange(x)) or any(not knows(i,cand) for i in xrange(n)):
            return -1
        return cand