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我们遍历长度为k的窗口#xff0c;用一个哈希表记录窗口内的所有元素#xff08;用来对窗口内元素去重#xff09;#xff0c;我们取哈希表中元素数量等于k的窗口总和的最大值
题解代码
use std::collecti… 文章目录 题目描述题解思路题解代码题目链接 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等于k的窗口总和的最大值
题解代码
use std::collections::{HashMap};impl Solution {pub fn maximum_subarray_sum(nums: Veci32, 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() k 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() k as usize {ans ans.max(sum);}}ans}
}题目链接
https://leetcode.cn/problems/maximum-sum-of-distinct-subarrays-with-length-k/description/
