关于二分查找

Posted 2021-09-04 Aiden_Zhao

二分查找

二分查找及其应用整理（假设数组有序）。

1. 查找与target相等的数字

def binarySearch(arr, target):
    left, right = 0, len(arr)-1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target: return mid
        elif arr[mid] > target: right = mid - 1
        else: left = mid + 1
    return -1

2. 查找最后一个小于target的数字

def binarySearch(arr, target):
    left, right = 0, len(arr)-1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] >= target: right = mid - 1
        else: left = mid + 1
    return right if right >= 0 else -1

3. 查找最后一个小于等于target的数字

def binarySearch(arr, target):
    left, right = 0, len(arr)-1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] > target: right = mid - 1
        else: left = mid + 1
    return right if right >= 0 else -1

4. 查找第一个大于target的数字

def binarySearch(arr, target):
    left, right = 0, len(arr)-1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] <= target: left = mid + 1
        else: right = mid - 1
    return left if left < len(arr) else -1

5. 查找第一个大于等于target的数字

def binarySearch(arr, target):
    left, right = 0, len(arr)-1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] < target: left = mid + 1
        else: right = mid - 1
    return left if left < len(arr) else -1

6. 查找最接近target的数字

def binarySearch(arr, target):
    left, right = 0, len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] <= target: left = mid + 1
        else: right = mid - 1
    if left >= len(arr): return len(arr) - 1
    return left if left == 0 or arr[left]-target <= target-arr[left-1] else left - 1