可能存在重复值的排序数组值

def binary_search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = left + (right - left) // 2
        if nums[mid] == target:
            # 处理重复值
            start, end = mid, mid
            while start > 0 and nums[start - 1] == target:
                start -= 1
            while end < len(nums) - 1 and nums[end + 1] == target:
                end += 1
            return (start, end)
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return None

# 示例数组
nums = [1, 2, 2, 2, 3, 4, 5]
target = 2
result = binary_search(nums, target)
print(result)  # 输出: (1, 3)