1695. Maximum Erasure Value

题目

You are given an array of positive integers nums and want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements.

Return the maximum score you can get by erasing exactly one subarray.

An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).

Example 1:

Input: nums = [4,2,4,5,6]
Output: 17
Explanation: The optimal subarray here is [2,4,5,6].

Example 2:

Input: nums = [5,2,1,2,5,2,1,2,5]
Output: 8
Explanation: The optimal subarray here is [5,2,1] or [1,2,5].

Constraints:

1 <= nums.length <= 105
1 <= nums[i] <= 104

题目大意

给你一个正整数数组 nums ，请你从中删除一个含有若干不同元素的子数组。删除子数组的得分就是子数组各元素之和。返回只删除一个子数组可获得的最大得分。如果数组 b 是数组 a 的一个连续子序列，即如果它等于 a[l],a[l+1],...,a[r] ，那么它就是 a 的一个子数组。

解题思路

读完题立马能识别出这是经典的滑动窗口题。利用滑动窗口从左往右滑动窗口，滑动过程中统计频次，如果是不同元素，右边界窗口又移，否则左边窗口缩小。每次移动更新 max 值。最终扫完一遍以后，max 值即为所求。

代码

package leetcode

func maximumUniqueSubarray(nums []int) int {
	if len(nums) == 0 {
		return 0
	}
	result, left, right, freq := 0, 0, -1, map[int]int{}
	for left < len(nums) {
		if right+1 < len(nums) && freq[nums[right+1]] == 0 {
			freq[nums[right+1]]++
			right++
		} else {
			freq[nums[left]]--
			left++
		}
		sum := 0
		for i := left; i <= right; i++ {
			sum += nums[i]
		}
		result = max(result, sum)
	}
	return result
}

func max(a int, b int) int {
	if a > b {
		return a
	}
	return b
}

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