2017-07-06 18:04
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slice1 := []string{"foo", "bar","hello"}
slice2 := []string{"foo", "bar"}

intersection(slice1, slice2)
=> ["foo", "bar"]
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5条回答 默认 最新

  • douzi2778 2017-07-06 18:59

    Yes there are a few different ways to go about it.. Here's an example that can be optimized.

    package main
    import "fmt"
    func intersection(a []string, b []string) (inter []string) {
        // interacting on the smallest list first can potentailly be faster...but not by much, worse case is the same
        low, high := a, b
        if len(a) > len(b) {
            low = b
            high = a
        done := false
        for i, l := range low {
            for j, h := range high {
                // get future index values
                f1 := i + 1
                f2 := j + 1
                if l == h {
                    inter = append(inter, h)
                    if f1 < len(low) && f2 < len(high) {
                        // if the future values aren't the same then that's the end of the intersection
                        if low[f1] != high[f2] {
                            done = true
                    // we don't want to interate on the entire list everytime, so remove the parts we already looped on will make it faster each pass
                    high = high[:j+copy(high[j:], high[j+1:])]
            // nothing in the future so we are done
            if done {
    func main() {
        slice1 := []string{"foo", "bar", "hello", "bar"}
        slice2 := []string{"foo", "bar"}
    ", intersection(slice1, slice2))

    Now the intersection method defined above will only operate on slices of strings, like your example.. You can in theory create a definition that looks like this func intersection(a []interface, b []interface) (inter []interface), however you would be relying on reflection and type casting so that you can compare, which will add latency and make your code harder to read. It's probably easier to maintain and read to write a separate function for each type you care about.

    func intersectionString(a []string, b []string) (inter []string),

    func intersectionInt(a []int, b []int) (inter []int),

    func intersectionFloat64(a []Float64, b []Float64) (inter []Float64), ..ect

    You can then create your own package and reuse once you settle how you want to implement it.

    package intersection
    func String(a []string, b []string) (inter []string)
    func Int(a []int, b []int) (inter []int)
    func Float64(a []Float64, b []Float64) (inter []Float64)
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  • douliang2935 2017-07-06 18:16

    How do I get the intersection between two arrays as a new array?

    • Simple Intersection: Compare each element in A to each in B (O(n^2))
    • Hash Intersection: Put them into a hash table (O(n))
    • Sorted Intersection: Sort A and do an optimized intersection (O(n*log(n)))

    All of which are implemented here


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  • dongshicuo4844 2017-07-11 02:18

    if there exists no blank in your []string, maybe you need this simple code:

    func filter(src []string) (res []string) {
        for _, s := range src {
            newStr := strings.Join(res, " ")
            if !strings.Contains(newStr, s) {
                res = append(res, s)
    func intersections(section1, section2 []string) (intersection []string) {
        str1 := strings.Join(filter(section1), " ")
        for _, s := range filter(section2) {
            if strings.Contains(str1, s) {
                intersection = append(intersection, s)
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  • duanmei1850 2018-06-05 07:24

    It's a best method for intersection two slice. Time complexity is too low.

    Time Complexity : O(m+n)

    m = length of first slice.

    n = length of second slice.

    func intersection(s1, s2 []string) (inter []string) {
        hash := make(map[string]bool)
        for _, e := range s1 {
            hash[e] = true
        for _, e := range s2 {
            // If elements present in the hashmap then append intersection list.
            if hash[e] {
                inter = append(inter, e)
        //Remove dups from slice.
        inter = removeDups(inter)
    //Remove dups from slice.
    func removeDups(elements []string)(nodups []string) {
        encountered := make(map[string]bool)
        for _, element := range elements {
            if !encountered[element] {
                nodups = append(nodups, element)
                encountered[element] = true
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  • duanaozhong0696 2019-02-07 23:52


    Another O(m+n) Time Complexity solution that uses a hashmap. It has two differences compared to the other solutions discussed here.

    • Passing the target slice as a parameter instead of new slice returned
    • Faster to use for commonly used types like string/int instead of reflection for all
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