关于进一步加强网站建设和广告平台对接

题目描述

题解思路

我们遍历长度为k的窗口，用一个哈希表记录窗口内的所有元素（用来对窗口内元素去重），我们取哈希表中元素数量大于等于m的窗口总和的最大值

题解代码

use std::collections::HashMap;impl Solution {pub fn max_sum(nums: Vec<i32>, m: i32, k: i32) -> i64 {let mut win = HashMap::new();let mut sum = 0;for i in 0..k as usize {win.insert(nums[i], i);sum += nums[i] as i64;}let mut ans = 0i64;if win.len() >= m as usize {ans = ans.max(sum);}for i in k as usize..nums.len() {if let Some(&start) = win.get(&nums[i-k as usize]) {if start == i - k as usize {win.remove(&nums[i-k as usize]);}}win.insert(nums[i], i);sum += (nums[i] - nums[i-k as usize]) as i64;if win.len() >= m as usize {ans = ans.max(sum);}}ans}
}

题目链接

https://leetcode.cn/problems/maximum-sum-of-almost-unique-subarray/description/