Huangzj
59 lines
1.6 KiB
Go
59 lines
1.6 KiB
Go
/*
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* @Author : huangzj
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* @Time : 2020/8/18 14:11
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* @Description：Lomuto划分,找到一个数组中第k个小的元素.这边只是提供一个示例，具体类型可以根据该算法进行修改
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*/
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package search
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var (
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times = 0 //用来计算一次划分进行了多少次的比较操作
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charge = 0 //实际进行元素交换的次数
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)
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//参数说明：
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//list ：数组
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//start : 开始比较的数字下标
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//end ：结束比较的数字下标
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//作用说明：
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func LomutoPartition(list []int, start int, end int) int {
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moveSubscript := start
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compareValue := list[start]
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for i := start + 1; i <= end; i++ {
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if list[i] < compareValue {
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moveSubscript++
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charge++
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list[moveSubscript], list[i] = list[i], list[moveSubscript] //交换元素
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}
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}
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charge++
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list[moveSubscript], list[start] = list[start], list[moveSubscript] //交换元素
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return moveSubscript
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}
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func LomutoQuiteSelect(list []int, pointPosition int) (int, int, int) {
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positionCheck(pointPosition, len(list)) //下标校验
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times = 0
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charge = 0
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nowPosition := LomutoPartition(list, 0, len(list)-1)
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times++
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//循环进行lomuto划分处理,知道找到我们需要的那个位置的元素
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for nowPosition != pointPosition-1 {
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times++
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if nowPosition > pointPosition-1 {
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nowPosition = LomutoPartition(list, 0, nowPosition)
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} else {
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nowPosition = LomutoPartition(list, nowPosition+1, len(list)-1)
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}
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}
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return list[nowPosition], times, charge
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}
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func positionCheck(pointPosition int, len int) {
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if pointPosition <= 0 || pointPosition > len {
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panic("查询的元素下标不能小于0，或者大于数组长度")
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}
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}
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