35. 搜索插入位置
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)
链接:https://leetcode-cn.com/problems/search-insert-position/
Link:https://leetcode.com/problems/search-insert-position/
暴力破解
O(N)
从头到尾扫描一遍,就拿到答案了。
二分法
O(logN)
由于数组中的元素是唯一的,所以遇到相等的就可以返回了
⚠️注意数组为空,或者数组元素都比target小的情况
模版一(left + 1 < right)
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
if len(nums) == 0:
return 0
low = 0
high = len(nums) - 1
while (low + 1 < high):
mid = low + (high - low) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
high = mid
else:
low = mid
if nums[low] >= target:
return low
elif nums[high] >= target:
return high
else:
return high + 1
模版二(left < right)
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
if len(nums) == 0:
return 0
low = 0
high = len(nums) - 1
while low < high:
mid = low + (high - low) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
high = mid
else:
low = mid + 1
if nums[low] >= target:
return low
else:
return low + 1
模版三(left <= right)
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
low = 0
high = len(nums) - 1
while low <= high:
mid = low + (high - low) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
high = mid - 1
else:
low = mid + 1
return low
–End–