leetcode

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Published: Dec 11, 2024 License: MIT Imports: 0 Imported by: 0

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1673. Find the Most Competitive Subsequence

题目

Given an integer array nums and a positive integer k, return the most competitive subsequence of nums of size k.

An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.

We define that a subsequence a is more competitive than a subsequence b (of the same length) if in the first position where a and b differ, subsequence a has a number less than the corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the first position they differ is at the final number, and 4 is less than 5.

Example 1:

Input: nums = [3,5,2,6], k = 2
Output: [2,6]
Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive.

Example 2:

Input: nums = [2,4,3,3,5,4,9,6], k = 4
Output: [2,3,3,4]

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
  • 1 <= k <= nums.length

题目大意

给你一个整数数组 nums 和一个正整数 k ,返回长度为 k 且最具 竞争力 的 nums 子序列。数组的子序列是从数组中删除一些元素(可能不删除元素)得到的序列。

在子序列 a 和子序列 b 第一个不相同的位置上,如果 a 中的数字小于 b 中对应的数字,那么我们称子序列 a 比子序列 b(相同长度下)更具 竞争力 。 例如,[1,3,4] 比 [1,3,5] 更具竞争力,在第一个不相同的位置,也就是最后一个位置上, 4 小于 5 。

解题思路

  • 这一题是单调栈的典型题型。利用单调栈,可以保证原数组中元素相对位置不变,这满足题意中删除元素但不移动元素的要求。单调栈又能保证每次进栈,元素是最小的。
  • 类似的题目还有第 42 题,第 84 题,第 496 题,第 503 题,第 856 题,第 901 题,第 907 题,第 1130 题,第 1425 题,第 1673 题。

代码

package leetcode

// 单调栈
func mostCompetitive(nums []int, k int) []int {
	stack := make([]int, 0, len(nums))
	for i := 0; i < len(nums); i++ {
		for len(stack)+len(nums)-i > k && len(stack) > 0 && nums[i] < stack[len(stack)-1] {
			stack = stack[:len(stack)-1]
		}
		stack = append(stack, nums[i])
	}
	return stack[:k]
}

Documentation

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