Python-二分查找算法框架

参考《力扣探索：二分查找》

def binarySearch(nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: int
    """
    if len(nums) == 0:
        return -1

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

    # End Condition: left > right
    return -1

它用于查找需要访问数组中当前索引及其直接右邻居索引的元素或条件**。

def binarySearch(nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: int
    """
    if len(nums) == 0:
        return -1

    left, right = 0, len(nums)
    while left < right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid

    # Post-processing:
    # End Condition: left == right
    if left != len(nums) and nums[left] == target:
        return left
    return -1

它用于搜索需要访问当前索引及其在数组中的直接左右邻居索引的元素或条件**。

def binarySearch(nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: int
    """
    if len(nums) == 0:
        return -1

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

    # Post-processing:
    # End Condition: left + 1 == right
    if nums[left] == target: return left
    if nums[right] == target: return right
    return -1

例题 Leetcode 875. 爱吃香蕉的珂珂
link：https://leetcode-cn.com/problems/koko-eating-bananas/

class Solution(object):
    def minEatingSpeed(self, piles, H):
        # Can Koko eat all bananas in H hours with eating speed K?
        def possible(K):
            return sum((p-1) / K + 1 for p in piles) <= H

        lo, hi = 1, max(piles)
        while lo < hi:
            mi = (lo + hi) / 2
            if not possible(mi):
                lo = mi + 1
            else:
                hi = mi
        return lo