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] + 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.
Solution
Counting sort, time O(n), space O(n)
原本用二分做的，后来发现还是counting sort更给力，果然一涉及到年龄什么的啊就一定要想到用Counting sort！
首先统计每个年龄对应的人个数，然后根据题目列出的条件来限定范围，在两个集合 i 和 j 中各选出一个做组合（单向request）即可。注意如果 i == j，则需要在 i 中选出两个元素做排列（双向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;
}
}
