277. Find the Celebrity (1)

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). 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.

Solution

已知celebtiy一定不認識其他任何人，然後其他人必定認識他。

先設置一個candidate指向第一個人(index = 0)，然後去判斷是否candidate有認識到人，若有，就不是celebrity，然後將認識的人指派t成新的candidate 。

找到最後可能的candidate。 要避免誤判，可能前第一輪迴圈有沒有確認到的人 i 認識candidate ( i < candidate)。所以要針對candidate再進行一次總檢查(其他人必認識candidate， candidate不認識其他人)。若不符合就是沒有celebrity。

class Solution {
    public int findCelebrity(int n) {
        int res = 0;
        for (int i = 0; i < n; ++i) {
            if (knows(res, i)) res = i;
        }
        for (int i = 0; i < n; ++i) {
            if (res != i && (knows(res, i) || !knows(i, res))) return -1;
        }
        return res;
    }
}