【模板题】二分法 - 34. 在排序数组中查找元素的第一个和最后一个位置

题目链接	34. 在排序数组中查找元素的第一个和最后一个位置
思路	二分法
题解链接	【视频讲解】二分查找总是写不对？三种写法，一个视频讲透！（Python/Java/C++/C/Go/JS）
关键点	模板题；应当熟练掌握
时间复杂度	$O(\log n)$
空间复杂度	$O(1)$

代码实现：

# 闭区间
def lower_bound(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        # 循环不变量：
        # nums[left-1] < target
        # nums[right+1] >= target
        mid = (left + right) // 2
        if nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return left
 
# 左闭右开区间
def lower_bound2(nums, target):
    left, right = 0, len(nums)
    while left < right:
        mid = (left + right) // 2
        # 循环不变量：
        # nums[left] < target
        # nums[right] >= target
        if nums[mid] < target:
            left = mid + 1
        else:
            right = mid
    return left
 
 
# 开区间
def lower_bound3(nums, target):
    left, right = -1, len(nums)
    while left + 1 < right:
        mid = (left + right) // 2
        # 循环不变量：
        # nums[left] < target
        # nums[right] >= target
        if nums[mid] < target:
            left = mid
        else:
            right = mid
    return right
 
 
class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        start = lower_bound(nums, target)
        if start == len(nums) or nums[start] != target:
            return [-1, -1]
        end = lower_bound3(nums, target+1) - 1
        return [start, end]

2024-09-12
重点是循环不变量：

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        n = len(nums)
 
        # start position
        # nums[left] < target && nums[right] >= target
        left, right = -1, n
        while left + 1 < right:
            mid = (left+right) // 2
            if nums[mid] < target:
                left = mid
            else:
                right = mid
        if right == n or nums[right] != target:
            return [-1, -1]
        start = right
        
        # end position
        # nums[left] <= target && nums[right] > target
        left, right = -1, n
        while left + 1 < right:
            mid = (left+right) // 2
            if nums[mid] <= target:
                left = mid
            else:
                right = mid
 
        end = right - 1
        return [start, end]