leetcode277- Find the Celebrity- medium

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.

 

算法：关键是每问一次问题必定可以淘汰一个人。时间复杂度是O(n)。第一次遍历循环问一个问题，可以只剩下一个人。第二次遍历循环问两个问题，可以确定这个人是不是真金。

1.空间复杂度O(n)：用Queue来存candidates，每次拉出来，重新推进去那个没被淘汰的。

2.空间复杂度O(1)：用一个int变量来存往前走一路一直只剩下的唯一的那个candidate。

细节：final是java关键字，命名的时候不能用，可以用last替代。

 

 

1.空间复杂度O(n): queue

/* 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) {
        if (n <= 1) {
            return -1;
        }
        Queue<Integer> candidates = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            candidates.offer(i);
        }
        
        while (candidates.size() > 1) {
            int a = candidates.poll();
            int b = candidates.poll();
            if (knows(a, b)) {
                candidates.offer(b);
            } else {
                candidates.offer(a);
            }
        }
        
        int last = candidates.poll();
        for (int i = 0; i < n; i++) {
            if (i == last) {
                continue;
            }
            if (knows(last, i) || !knows(i, last)) {
                return -1;
            }
        }
        return last;
    }
}

 

2.空间复杂度O(1)

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// version: 高频题班
public class Solution extends Relation {
    /**
     * @param n a party with n people
     * @return the celebrity's label or -1
     */
    public int findCelebrity(int n) {
        // Write your code here
        int ans = 0;
        for (int i = 1; i < n; i++) {
            if (knows(ans, i)) {
                ans = i;
            }
        }

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