leetcode 277. Find the Celebrity【思维】

277. Find the Celebrity

Medium

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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 - 1people 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.

要找出一个leader 使得别人所有人都认识他 他谁都不认识

做法很巧妙：

先假定一个人是 挨个枚举其他人看他是否认识i 发现candidate认识i candidate换成i

再遍历一遍，看找到的这个对不对

我前几天一直纠结

为啥这个会不漏下正解呢

因为正解只会有一个 

如果 第一次循环找到了这样的一个数，如果有多个满足，第二次for循环就会pass掉

// Forward declaration of the knows API.
bool knows(int a, int b);

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