Description

Some people will make friend requests. The list of their ages is given and ages[i] is the age of the ith person.

Person A will NOT friend request person B (B != A) if any of the following conditions are true:

• age[B] <= 0.5 * age[A] + 7
• age[B] > age[A]
• age[B] > 100 && age[A] < 100

Otherwise, A will friend request B.

Note that if A requests B, B does not necessarily request A. Also, people will not friend request themselves.

How many total friend requests are made?

Example 1:

Input: [16,16]
Output: 2
Explanation: 2 people friend request each other.

Example 2:

Input: [16,17,18]
Output: 2
Explanation: Friend requests are made 17 -> 16, 18 -> 17.

Example 3:

Input: [20,30,100,110,120]
Output:
Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.

Notes:

• 1 <= ages.length <= 20000.
• 1 <= ages[i] <= 120.

Solution

Counting sort, time O(n), space O(n)

原本用二分做的，后來發(fā)現(xiàn)還是counting sort更給力，果然一涉及到年齡什么的啊就一定要想到用Counting sort！

首先統(tǒng)計(jì)每個(gè)年齡對應(yīng)的人個(gè)數(shù)，然后根據(jù)題目列出的條件來限定范圍，在兩個(gè)集合 i 和 j 中各選出一個(gè)做組合（單向request）即可。注意如果 i == j，則需要在 i 中選出兩個(gè)元素做排列（雙向request）！

class Solution {
    public int numFriendRequests(int[] ages) {
        int n = ages.length;
        int[] count = new int[121];
        for (int age : ages) {
            ++count[age];
        }
        
        int requests = 0;
        for (int i = 1; i <= 120; ++i) {
            for (int j = i / 2 + 8; j <= i; ++j) {
                if (j == i) {
                    requests += count[i] * (count[i] - 1);  // watch out
                } else {
                    requests += count[i] * count[j];
                }
            }
        }
        
        return requests;
    }
}