二分查找及几种变体的Python实现

1. 在不重复的有序数组中，查找等于给定值的元素

循环法

def search(lst, target):
    n = len(lst)
    if n == 0:
        return -1
    
    low = 0
    high = n - 1
    while low <= high:
        mid = (high-low)//2 + low
        if lst[mid] == target:
            return mid
        elif lst[mid] < target:
            low = mid +1
        elif lst[mid] > target:
            high = mid -1
    return -1

递归法

def search2(lst, target, low, high):
    if low > high:
        return -1
    
    mid = (high-low)//2 + low
    if lst[mid] == target:
        return mid
    elif lst[mid] < target:
        return search2(lst, target, mid+1, high)
    elif lst[mid] > target:
        return search2(lst, target, low, mid-1)

2. 查找第一个值等于给定值的元素

def bsearch(lst, value):
    n = len(lst)
    if n == 0:
        return -1
    low = 0
    high = n - 1
    while low <= high:
        mid = (high-low) // 2 + low
        if lst[mid] < value:
            low = mid + 1
        elif lst[mid] > value:
            low = mid - 1
        elif lst[mid] == value:
            if mid == 0 or lst[mid-1] != value:
                return mid
            else:
                high = mid - 1
    return -1

3. 查找最后一个值等于给定值的元素

def bsearch(lst, value):
    n = len(lst)
    if n == 0:
        return -1
    low = 0
    high = n - 1
    while low <= high:
        mid = (high-low) // 2 + low
        if lst[mid] > value:
            high = mid - 1
        elif lst[mid] < value:
            low = mid + 1
        elif lst[mid] == value:
            if mid == n-1 or lst[mid+1] != value:
                return mid
            else:
                low = mid + 1
    return -1

4. 查找第一个大于等于给定值的元素

def bsearch(lst, value):
    n = len(lst)
    if n == 0:
        return -1
    low = 0
    high = n - 1
    while low <= high:
        mid = (high-low) // 2 + low
        if lst[mid] < value:
            low = mid + 1
        elif lst[mid] >=value:
            if mid == 0 or lst[mid-1] < value:
                return mid
            high = mid - 1

5. 查找最后一个小于等于给定值的元素

def bsearch(lst, value):
    n = len(lst)
    if n == 0:
        return -1
    low = 0
    high = n - 1
    while low <= high:
        mid = (high-low) // 2 + low
        if lst[mid] > value:
            high = mid - 1
        elif lst[mid] <= value:
            if mid == n-1 or lst[mid+1] > 0:
                return mid
            low = mid + 1
    return -1