277. Find the Celebrity

题目：

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

.

链接： http://leetcode.com/problems/find-the-celebrity/

题解：

在数组中找Celebrity。Celebrity的条件很特殊，首先所有人都认识Celebrity，其次Celebrity不认识任何人。一开始想的方法是Brute force，需要O(n2)遍历。看了Discuss以后发现我们需要好好利用第一个条件，就是假定candidate = 0，遍历数组，当knows(candidate, i)时， 假定i为candidate。这里因为所有人都认识Celebrity，所以遍历完数组以后candidate就是我们要验证的Celebrity。但也有还需要第二遍遍历来验证这个candidate是否满足我们的条件。

Time Complexity - O(n)， Space Complexity - O(1)

/* 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 candidate = 0;
for(int i = 1; i < n; i++) {
if(knows(candidate, i)) {
candidate = i;
}
}

for(int i = 0; i < n; i++) {
if(i != candidate && (knows(candidate, i) || !knows(i, candidate))) {
return -1;
}
}

return candidate;
}
}

Reference:
https://leetcode.com/discuss/56350/straight-forward-c-solution-with-explaination https://leetcode.com/discuss/56413/java-solution-two-pass https://leetcode.com/discuss/61140/java-python-o-n-calls-o-1-space-easy-to-understand-solution https://leetcode.com/discuss/56349/python-brute-force-solution-easy-understanding