算法:二分查找

算法 About 2,085 words

循环法

func main() {
    arr := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9}

    fmt.Println(BinarySearchByRecursion(arr, 0, len(arr)-1, 9))
    //l := BinarySearchByLoop(arr, 9)
    //fmt.Printf("%+v", l)
}

func BinarySearchByLoop(arr []int, no int) int {
    left := 0
    right := len(arr) - 1
    for left <= right {
        mid := (left + right) / 2
        if no == arr[mid] {
            fmt.Println("找到了,索引#", mid)
            return mid
        }
        if no < arr[mid] {
            right = mid - 1
        }
        if no > arr[mid] {
            left = mid + 1
        }
    }
    return -1
}

递归法

func BinarySearchByRecursion(arr []int, left, right, no int) int {
    if left > right {
        return -1
    }
    mid := (left + right) / 2

    if no < arr[mid] {
        return BinarySearchByRecursion(arr, left, right-1, no)
    }

    if no > arr[mid] {
        return BinarySearchByRecursion(arr, left+1, right, no)
    }

    // 往左找
    temp := mid - 1
    for temp >= left && arr[temp] == no {
        fmt.Println("往左找找到一个#", temp)
        temp--
    }
    // 往右找
    temp = mid + 1
    for temp <= right && arr[temp] == no {
        fmt.Println("往右找找到一个#", temp)
        temp++
    }
    return mid
}

循环法-查询多个值

func BinarySearchByLoop(arr []int, no int) *list.List {
    l := list.New()
    left := 0
    right := len(arr) - 1
    for left <= right {
        mid := (left + right) / 2
        if no == arr[mid] {
            fmt.Println("找到了,索引#", mid)
            l.PushBack(mid)
            // 往左找
            temp := mid - 1
            for temp >= left && arr[temp] == no {
                l.PushBack(temp)
                temp--
            }
            // 往右找
            temp = mid + 1
            for temp <= right && arr[temp] == no {
                l.PushBack(temp)
                temp++
            }
            return l
        }
        if no < arr[mid] {
            right = mid - 1
        }
        if no > arr[mid] {
            left = mid + 1
        }
    }
    return nil
}
Views: 1,890 · Posted: 2021-02-10

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