class Solution {
public:
int arithmeticTriplets(vector<int>& nums, int diff) {
}
};
2367. 算术三元组的数目
给你一个下标从 0 开始、严格递增 的整数数组 nums 和一个正整数 diff 。如果满足下述全部条件，则三元组 (i, j, k) 就是一个 算术三元组 ：
i < j < k ，nums[j] - nums[i] == diff 且nums[k] - nums[j] == diff返回不同 算术三元组 的数目。
示例 1：
输入：nums = [0,1,4,6,7,10], diff = 3 输出：2 解释： (1, 2, 4) 是算术三元组：7 - 4 == 3 且 4 - 1 == 3 。 (2, 4, 5) 是算术三元组：10 - 7 == 3 且 7 - 4 == 3 。
示例 2：
输入：nums = [4,5,6,7,8,9], diff = 2 输出：2 解释： (0, 2, 4) 是算术三元组：8 - 6 == 2 且 6 - 4 == 2 。 (1, 3, 5) 是算术三元组：9 - 7 == 2 且 7 - 5 == 2 。
提示：
3 <= nums.length <= 2000 <= nums[i] <= 2001 <= diff <= 50nums 严格 递增原站题解
swift 解法, 执行用时: 8 ms, 内存消耗: 13.8 MB, 提交时间: 2023-09-24 23:42:27
class Solution {
func arithmeticTriplets(_ nums: [Int], _ diff: Int) -> Int {
// 定义出现次数变量, 定义哈希表temp 添加nums前2个元素,
// temp初始为 [nums[0] : 1, nums[1] : 1]
var res = 0, temp = [nums[0] : 1, nums[1] : 1]
// 遍历, 因为前2个元素已经加入
for i in 2..<nums.count {
// 如果哈希表存在 nums[i] - diff, nums[i] - diff * 2
if temp[nums[i] - diff] != nil && temp[nums[i] - diff * 2] != nil {
// 结果 + 1
res += 1
}
// 新增键值对, 当前数字做`Key`, 1为`value`
temp[nums[i]] = 1
}
// 返回结果
return res
}
}
scala 解法, 执行用时: 564 ms, 内存消耗: 55.2 MB, 提交时间: 2023-09-24 23:40:50
object Solution {
def arithmeticTriplets(nums: Array[Int], diff: Int): Int = {
var right = nums.groupBy(x=>x).mapValues(_.length).toMap;
var left = Map[Int, Int]()
var ans = 0
for (i <- nums){
right = right + (i -> (right.getOrElse(i, 0) - 1))
ans += left.getOrElse(i-diff, 0) * right.getOrElse(i+diff, 0)
left = left + (i -> (left.getOrElse(i, 0) + 1))
}
ans
}
}
rust 解法, 执行用时: 0 ms, 内存消耗: 2.2 MB, 提交时间: 2023-09-12 17:53:00
// hash
impl Solution {
pub fn arithmetic_triplets(nums: Vec<i32>, diff: i32) -> i32 {
use std::collections::HashSet;
let (mut cnt, mut cache) = (0, HashSet::new());
nums.into_iter().for_each(|num| {
if cache.contains(&(num - diff)) && cache.contains(&(num - 2 * diff)) { cnt += 1 };
cache.insert(num);
});
cnt
}
}
/* 三指针
impl Solution {
pub fn arithmetic_triplets(nums: Vec<i32>, diff: i32) -> i32 {
let (mut i, mut j, mut k, mut cnt) = (0, 1, 2, 0);
while k < nums.len() {
while j < k && nums[j] + diff < nums[k] { j += 1; }
if nums[j] + diff == nums[k] {
while i < j && nums[i] + diff < nums[j] { i += 1; }
if nums[i] + diff == nums[j] { cnt += 1; }
}
k += 1;
}
cnt
}
}
*/
php 解法, 执行用时: 8 ms, 内存消耗: 18.8 MB, 提交时间: 2023-09-12 17:51:39
class Solution {
/**
* @param Integer[] $nums
* @param Integer $diff
* @return Integer
*/
function arithmeticTriplets($nums, $diff) {
$map = [];
foreach ($nums as $num) {
$map[$num] = 1;
}
$ans = 0;
foreach ($nums as $num) {
if (
isset($map[$num + $diff]) &&
isset($map[$num - $diff])
)
$ans++;
}
return $ans;
}
}
golang 解法, 执行用时: 0 ms, 内存消耗: 2.1 MB, 提交时间: 2022-08-12 11:37:49
func arithmeticTriplets(nums []int, diff int) (ans int) {
i, j := 0, 1
for _, x := range nums[2:] {
for nums[j]+diff < x {
j++
}
if nums[j]+diff > x {
continue
}
for nums[i]+diff*2 < x {
i++
}
if nums[i]+diff*2 == x {
ans++
}
}
return
}
golang 解法, 执行用时: 0 ms, 内存消耗: 2.3 MB, 提交时间: 2022-08-12 11:37:26
func arithmeticTriplets(nums []int, diff int) (ans int) {
set := map[int]bool{}
for _, x := range nums {
if set[x-diff] && set[x-diff*2] {
ans++
}
set[x] = true
}
return
}
golang 解法, 执行用时: 0 ms, 内存消耗: 2.3 MB, 提交时间: 2022-08-12 11:36:34
func arithmeticTriplets(nums []int, diff int) (ans int) {
set := map[int]bool{}
for _, x := range nums {
set[x] = true
}
for _, x := range nums {
if set[x-diff] && set[x+diff] {
ans++
}
}
return
}
python3 解法, 执行用时: 32 ms, 内存消耗: 14.8 MB, 提交时间: 2022-08-12 11:36:07
class Solution:
def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
s = set(nums)
return sum(x - diff in s and x + diff in s for x in nums)