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| // @Title: 数组中的第K个最大元素 (Kth Largest Element in an Array)
// @Author: 15816537946@163.com
// @Date: 2019-11-17 09:38:56
// @Runtime: 12 ms
// @Memory: 3.5 MB
/*
* @lc app=leetcode.cn id=215 lang=golang
*
* [215] 数组中的第K个最大元素
*
* https://leetcode-cn.com/problems/kth-largest-element-in-an-array/description/
*
* algorithms
* Medium (58.03%)
* Likes: 240
* Dislikes: 0
* Total Accepted: 40.5K
* Total Submissions: 69.7K
* Testcase Example: '[3,2,1,5,6,4]\n2'
*
* 在未排序的数组中找到第 k 个最大的元素。请注意,你需要找的是数组排序后的第 k 个最大的元素,而不是第 k 个不同的元素。
*
* 示例 1:
*
* 输入: [3,2,1,5,6,4] 和 k = 2
* 输出: 5
*
*
* 示例 2:
*
* 输入: [3,2,3,1,2,4,5,5,6] 和 k = 4
* 输出: 4
*
* 说明:
*
* 你可以假设 k 总是有效的,且 1 ≤ k ≤ 数组的长度。
*
*/
// quick sort
func findKthLargest(nums []int, k int) int {
return doFindKthLargest(nums, k,0, len(nums)-1)
}
func doFindKthLargest(nums []int, k, start,end int) int {
nLen := len(nums)
targetPos := nLen-k
for {
povitIndex := partition(nums, start, end)
if povitIndex == targetPos {
return nums[povitIndex]
} else if povitIndex > targetPos {
end= povitIndex-1
} else {
start = povitIndex+1
}
}
}
func partition(nums []int, start, end int) int {
povit := nums[start]
lo,hi := start+1, end
for lo<=hi{
for lo <= hi && nums[lo]<=povit {
lo++
}
for lo <= hi && nums[hi] > povit {
hi--
}
if lo< hi {
nums[lo],nums[hi] = nums[hi],nums[lo]
}
}
nums[start], nums[hi] = nums[hi], nums[start]
return hi
}
|