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141 lines
4.6 KiB
Markdown
141 lines
4.6 KiB
Markdown
# [825. Friends Of Appropriate Ages](https://leetcode.com/problems/friends-of-appropriate-ages/)
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## 题目
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There are `n` persons on a social media website. You are given an integer array `ages` where `ages[i]` is the age of the `ith` person.
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A Person `x` will not send a friend request to a person `y` (`x != y`) if any of the following conditions is true:
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- `age[y] <= 0.5 * age+ 7`
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- `age[y] > age[x]`
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- `age[y] > 100 && age< 100`
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Otherwise, `x` will send a friend request to `y`.
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Note that if `x` sends a request to `y`, `y` will not necessarily send a request to `x`. Also, a person will not send a friend request to themself.
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Return *the total number of friend requests made*.
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**Example 1:**
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```
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Input: ages = [16,16]
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Output: 2
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Explanation: 2 people friend request each other.
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```
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**Example 2:**
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```
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Input: ages = [16,17,18]
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Output: 2
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Explanation: Friend requests are made 17 -> 16, 18 -> 17.
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```
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**Example 3:**
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```
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Input: ages = [20,30,100,110,120]
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Output: 3
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Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.
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```
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**Constraints:**
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- `n == ages.length`
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- `1 <= n <= 2 * 10^4`
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- `1 <= ages[i] <= 120`
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## 题目大意
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在社交媒体网站上有 n 个用户。给你一个整数数组 ages ，其中 ages[i] 是第 i 个用户的年龄。
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如果下述任意一个条件为真，那么用户 x 将不会向用户 y（x != y）发送好友请求：
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- ages[y] <= 0.5 * ages[x] + 7
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- ages[y] > ages[x]
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- ages[y] > 100 && ages[x] < 100
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否则，x 将会向 y 发送一条好友请求。注意，如果 x 向 y 发送一条好友请求，y 不必也向 x 发送一条好友请求。另外，用户不会向自己发送好友请求。返回在该社交媒体网站上产生的好友请求总数。
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## 解题思路
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- 解法三，暴力解法。先统计 [1,120] 范围内每个年龄的人数。然后利用题目中的三个判断条件，筛选符合条件的用户对。需要注意的是，相同年龄的人可以相互发送好友请求。不同年龄的人发送好友请求是单向的，即年龄老的向年龄轻的发送好友请求，年龄轻的不会对年龄老的发送好友请求。
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- 解法二，排序 + 双指针。题目给定的 3 个条件其实是 2 个。条件 3 包含在条件 2 中。条件 1 和条件 2 组合起来是 `0.5 × ages[x]+7 < ages[y] ≤ ages[x]`。当 ages[x] 小于 15 时，这个等式无解。考虑到年龄是单调递增的，`(0.5 × ages[x]+7,ages[x]]` 这个区间左右边界也是单调递增的。于是可以用双指针维护两个边界。在区间 [left, right] 内，这些下标对应的的 y 值都满足条件。当 `ages[left] > 0.5 × ages[x]+7` 时，左指针停止右移。当 `ages[right+1] > ages[x]` 时， 右指针停止右移。在 `[left, right]` 区间内，满足条件的 y 有 `right-left+1` 个，即使得 `ages[y]` 取值在 `(0.5 × ages[x]+7,ages[x]]` 之间。依照题意，`x≠y`，即该区间右边界取不到。y 的取值个数需要再减一，减去的是取到和 x 相同的值的下标。那么每个区间能取 `right-left` 个值。累加所有满足条件的值即为好友请求总数。
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- 解法一。在解法二中，计算满足不等式 y 下标所在区间的时候，区间和区间存在重叠的情况，这些重叠情况导致了重复计算。所以这里可以优化。可以用 prefix sum 前缀和数组优化。代码见下方。
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## 代码
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```go
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package leetcocde
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import "sort"
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// 解法一 前缀和，时间复杂度 O(n)
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func numFriendRequests(ages []int) int {
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count, prefixSum, res := make([]int, 121), make([]int, 121), 0
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for _, age := range ages {
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count[age]++
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}
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for i := 1; i < 121; i++ {
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prefixSum[i] = prefixSum[i-1] + count[i]
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}
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for i := 15; i < 121; i++ {
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if count[i] > 0 {
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bound := i/2 + 8
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res += count[i] * (prefixSum[i] - prefixSum[bound-1] - 1)
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}
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}
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return res
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}
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// 解法二 双指针 + 排序，时间复杂度 O(n logn)
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func numFriendRequests1(ages []int) int {
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sort.Ints(ages)
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left, right, res := 0, 0, 0
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for _, age := range ages {
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if age < 15 {
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continue
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}
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for ages[left]*2 <= age+14 {
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left++
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}
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for right+1 < len(ages) && ages[right+1] <= age {
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right++
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}
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res += right - left
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}
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return res
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}
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// 解法三 暴力解法 O(n^2)
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func numFriendRequests2(ages []int) int {
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res, count := 0, [125]int{}
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for _, x := range ages {
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count[x]++
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}
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for i := 1; i <= 120; i++ {
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for j := 1; j <= 120; j++ {
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if j > i {
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continue
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}
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if (j-7)*2 <= i {
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continue
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}
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if j > 100 && i < 100 {
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continue
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}
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if i != j {
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res += count[i] * count[j]
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} else {
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res += count[i] * (count[j] - 1)
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}
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}
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}
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return res
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}
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``` |