2021-08-17 线性查找与二分查找

原创于 2021-08-17 20:17:44 发布 · 164 阅读

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本文探讨了Python中线性搜索与二分搜索的实现，通过实例展示了如何在大量数据中快速定位50000000元素。重点介绍了递归版二分搜索算法，并分析了它们在效率上的区别。

def liner_search(li, val):
    for ind, v in enumerate(li):
        if v == val:
            return ind
    else:
        return None


def binary_search(li, val):
    left = 0
    right = len(li) - 1
    while left <= right:  # 候选区有值
        mid = (left + right) // 2  # 整除
        if li[mid] == val:
            return mid
        elif li[mid] > val:  # 说明查找值在左侧
            right = mid - 1
        else:  # 说明查找值在右侧时
            left = mid + 1
    return None


def binary_search2(low, hight, item, list):  # python 递归版
    if low > hight:
        return -1
    mid = low + (hight - low) // 2  # 计算mid值，注意是 hight-low
    if list[mid] == item:
        return mid
    if list[mid] < item:
        return binary_search2(mid + 1, hight, item, list)
    if list[mid] > item:
        return binary_search2(low, mid - 1, item, list)


li2 = list(range(100000000))
print(binary_search(li2, 50000000))
print(liner_search(li2, 50000000))
print(binary_search2(0, len(li2) - 1, 50000000, li2))

2 条评论

weiabc 2021.08.17
[code=csharp] #include <stdio.h> //线性查找 int LinearSearch(int arr[], int n, int val) { for (int i = 0; i < n; i++) { if (arr[i] == val) { return i; } } return -1; } //二分查找法，前提是元素已经有序 int BinarySearch(int arr[], int n, int val) { int left = 0, right = n - 1; while (left <= right) { int mid = (left + right) / 2; //这一句要在循环内，不然循环的条件(即查找的范围)得不到收敛 if (arr[mid] < val) //待查找值在右侧 { left = mid + 1; //待查数组左边界进行修正 } else if (arr[mid] == val) //待查找值已经找到 { return mid; //返回索引 } else //待查找值在左侧 { right = mid - 1; } } } int main(int argc, char const *argv[]) { int arr[] = { 1,2,3,4,5,10,12,16,19 }; int n = 9; //printf("%d", LinearSearch(arr, n, 5)); printf("%d", BinarySearch(arr, n, 16)); return 0; } [/code]
- weiabc回复weiabc 2021.08.17
  [code=objc] //二分查找法，前提是元素已经有序,递归版本 int BinarySearchR(int arr[], int n, int val, int left, int right) { int mid = (left + right) / 2; //这一句要在循环内，不然循环的条件(即查找的范围)得不到收敛 if (arr[mid] < val) //待查找值在右侧 { left = mid + 1; //待查数组左边界进行修正 BinarySearchR(arr, n, val, left, right); } else if (arr[mid] == val) //待查找值已经找到 { return mid; //返回索引 } else //待查找值在左侧 { right = mid - 1; BinarySearchR(arr, n, val, left, right); } } [/code]