线性复杂度选出第k小元素、中位数、最小的k个元素(golang实现)

Tue Sep 5, 2017

400 Words|Read in about 1 Min
Tags: 算法  

封装成函数:

//选出第k小元素,k为1~len(s)
func SelectKthMin(s []int, k int) int {
	k--
	lo, hi := 0, len(s)-1
	for {
		j := partition(s, lo, hi)
		if j < k {
			lo = j + 1
		} else if j > k {
			hi = j - 1
		} else {
			return s[k]
		}
	}
}
 
//选出中位数(比一半的元素小,比另一半的大)
func SelectMid(s []int) int {
	return SelectKthMin(s, len(s)/2)
}
 
//选出k个最小元素,k为1~len(s)
func SelectKMin(s []int, k int) []int {
	lo, hi := 0, len(s)-1
	for {
		j := partition(s, lo, hi)
		if j < k {
			lo = j + 1
		} else if j > k {
			hi = j - 1
		} else {
			return s[:k]
		}
	}
}
 
func partition(s []int, lo, hi int) int {
	i, j := lo, hi+1
	for {
		for {
			i++
			if i == hi || s[i] > s[lo] {
				break
			}
		}
		for {
			j--
			if j == lo || s[j] <= s[lo] {
				break
			}
		}
		if i >= j {
			break
		}
		swap(s, i, j)
	}
	swap(s, lo, j)
	return j
}
 
func swap(s []int, i int, j int) {
	s[i], s[j] = s[j], s[i]
}

测试:

s := []int{9, 0, 6, 5, 8, 2, 1, 7, 4, 3}
fmt.Println(SelectKthMin(s,1)) //第1小元素:0
fmt.Println(SelectMid(s)) //中位数:4
fmt.Println(SelectKMin(s,5)) //最小的5个数:0~4

输出:
0
4
[0 1 2 3 4]

See Also

Tue Sep 5, 2017

400 Words|Read in about 1 Min
Tags: 算法