两个切片内容相减的几种方法

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问题

已知：
	var a []string = []string{"11", "33", "22", "44"}
	var b []string = []string{"11", "22"}
	var b []string

预期：
	c = []string{"11", "22"}

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# 1，运维的思路

# 方法一

package main

import "fmt"

func main() {
	var (
		weightInstanceList []string = []string{"11", "22", "33", "44"}
		RelateInstance     []string = []string{"11", "22"}
		a                  []string // 期望：22,33,44
	)

	fmt.Println(weightInstanceList, RelateInstance)

	// 循环老的
	for _, instance := range weightInstanceList {
		// 传入待删除IP切片 老的IP 进行判断 是否相等 相等表示待删除 反之加入更新jenkinsFile列表
		isOk := IsContain(RelateInstance, instance)
		if !isOk {
			a = append(a, instance)
		}

	}

	fmt.Println("最后结果：", a) // a=["11","11","11"]

}

func IsContain(items []string, item string) bool {
	for _, eachItem := range items {
		if eachItem == item {
			return true
		}
	}
	return false
}

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# 方法二

package main

import "fmt"

func main() {
	var (
		weightInstanceList []string = []string{"11", "33", "22", "44"}
		RelateInstance     []string = []string{"11", "22"}
	)
	fmt.Println(weightInstanceList[1:])
	for _, instance := range weightInstanceList {
		for _, item := range RelateInstance {
			if instance == item {
				weightInstanceList = RemoveParam(weightInstanceList, instance)
			}

		}
	}
	fmt.Println("最后结果：", weightInstanceList) // a=["11","11","11"]
}

func RemoveParam(sli []string, n string) []string {
	for i := 0; i < len(sli); i++ {
		if sli[i] == n {
			if i == 0 {
				sli = sli[1:]
			} else if i == len(sli)-1 {
				sli = sli[:i]
			} else {
				sli = append(sli[:i], sli[i+1:]...)
			}
			i-- // 如果索引i被去掉后，原来索引i+1的会移动到索引i
		}
	}
	return sli
}

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# 2，开发的思路

# 方法一

package main

import "fmt"

func main() {
	var (
		weightInstanceList []string = []string{"11", "33", "22", "44"}
		RelateInstance     []string = []string{"11", "22"}
	)
	fmt.Println(diff(weightInstanceList, RelateInstance))
}
func diff(a, b []string) []string {
	var (
		r []string
		m = make(map[string]struct{}, len(a))
	)

	for _, v := range b {
		m[v] = struct{}{}
	}

	for _, v := range a {
		if _, ok := m[v]; !ok {
			r = append(r, v)
		}
	}

	return r
}

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# 方法二

package main

import "fmt"

func main() {
	var (
		weightInstanceList []string = []string{"11", "33", "22", "44"}
		RelateInstance     []string = []string{"11", "22"}
	)
	fmt.Println(diff(weightInstanceList, RelateInstance))
}
func diff(a, b []string) []string {
	var (
		r []string
		m = make(map[string]struct{}, len(a))
	)

	for _, v := range a {
		m[v] = struct{}{}
	}

	for _, v := range b {
		delete(m, v)
	}

	for k, _ := range m {
		r = append(r, k)
	}

	return r
}

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可以看出，运维处理该问题时思路会困在当前类型中予以实现，不如开发对数据结构以及特性掌握的牢固，运用起来也就不够灵活了。

上次更新: 2024/11/19, 23:11:42

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