leetcode35. 搜索插入位置 🌟
题目:
给定一个排序数组和一个目标值,在数组中找到目标值,并返回其索引。如果目标值不存在于数组中,返回它将会被按顺序插入的位置。
你可以假设数组中无重复元素。
示例 1:
输入: [1,3,5,6], 5
输出: 2
示例 2:
输入: [1,3,5,6], 2
输出: 1
示例 3:
输入: [1,3,5,6], 7
输出: 4
示例 4:
输入: [1,3,5,6], 0
输出: 0
来源:力扣(LeetCode)
解答:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution: def searchInsert(self, nums: List[int], target: int) -> int: def binarySearch(nums, target): low = 0 high = len(nums) - 1 while low <= high: cur = int((low + high) / 2) if nums[cur] == target: return cur, None elif nums[cur] > target: high = cur - 1 elif nums[cur] < target: low = cur + 1 return -1, low target, low = binarySearch(nums, target) return low if target == -1 else target
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
def binarySearch(nums, target, low, high): if low > high: return low cur = (low + high) // 2 if nums[cur] == target: return cur elif nums[cur] > target: return binarySearch(nums, target, low, cur - 1) else: return binarySearch(nums, target, cur + 1, high) return binarySearch(nums, target, 0, len(nums) - 1)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution: def searchInsert(self, nums: List[int], target: int) -> int: for i, num in enumerate(nums): if num >= target: return i return len(nums)