825. Friends Of Appropriate Ages
問題描述：
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] + 7age[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.
解題思路：
一開始想到的是暴力破解法：即O(n2)的方法，但是顯然OJ是不會讓你過的：）
這裏參考了votrubac 的解法
只能說我沒有好好分析題意
遇見這種不等式沒能去化簡
由條件一： age[B] <= 0.5 * age[A] + 7
和條件二： age[B] > age[A]
可得 age[A] < 0.5 * age[A] + 7
即 age[A] < 14
當年齡小於14時，是不可以發出好友請求的。
由於題目限定： 1 <= ages[i] <= 120.
我們可以將每個年齡出現的個數存入一個固定大小的數組（這將花費我們O(n)的時間）
而由於遍歷固定大小的數組的時間是一定的，實際上是O(1)
根據我們上面得出的條件：
我們將從年齡為15的開始向後遍歷
對當前年齡，我們又可以分為兩種情況：
1. 同齡人：cnt[i] * (cnt[i] - 1)
2. 比TA小且滿足要求的人：由j = 0.5 * i + 8開始向後遍歷，且j < i
代碼：
class Solution { public: int numFriendRequests(vector<int>& ages) { int cnt[121] = {}; int ret = 0; for(int a : ages){ cnt[a]++; } for(int i = 15; i < 121; i++){ if(cnt[i] == 0) continue; ret += cnt[i] * (cnt[i]-1); for(int j = 0.5 * i + 8; j < i; j++) ret += cnt[j]*cnt[i]; } return ret; } };
825. Friends Of Appropriate Ages