LintCode 1393: Friends Of Appropriate Ages-优快云博客

本文探讨了在社交网络中，基于年龄限制的朋友请求算法。通过分析不同年龄组的人是否可以互相发送朋友请求的规则，提出了两种算法实现，一种时间复杂度为O(n^2)，另一种更优的时间复杂度为O(K^2)，K为独特年龄的数量。

Friends Of Appropriate Ages

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
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: 3
Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.
Notice
1 <= ages.length <= 20000.
1 <= ages[i] <= 120.

解法1：
注意这题的两个条件
age[B] > age[A]
age[B] > 100 && age[A] < 100
有冗余。用age[B] > age[A]一个条件就可以了。

我想的解法比较慢。时间复杂度O(n^2)。刚刚勉强过。

class Solution {
public:
    /**
     * @param ages: 
     * @return: nothing
     */
    int numFriendRequests(vector<int> &ages) {
        int n = ages.size();
        sort(ages.begin(), ages.end());
        int numFriends = 0;
        
        for (int i = n - 1; i > 0; i--) {
            int A = ages[i];
            for (int j = i - 1; j >= 0; j--) {
                int B = ages[j];
                if (B <= A / 2 + 7) break;
                numFriends++;
                if (A == B) numFriends++;
            }
        }
        
        return numFriends;
        
    }
};

解法2：
参考的网上的答案。时间复杂度仅为O(K^2)，K为unique的年龄个数。

class Solution {
public:
    /**
     * @param ages: 
     * @return: nothing
     */
    int numFriendRequests(vector<int>& ages) {
        unordered_map<int,int> mp;   //age, freq
        for (int &age: ages) mp[age]++;
        int res = 0;
        for (auto & a: mp) 
           for (auto & b: mp)
            if (request(a.first, b.first)) res += a.second * (b.second - (a.first == b.first ? 1 : 0));
        return res;
    }

    bool request(int a, int b) {
        //return !(b <= 0.5 * a + 7 || b > a || (b > 100 && a < 100));
        return !(b <= 0.5 * a + 7 || b > a);
    }
};