34. Find First and Last Position of Element in Sorted Array

题目

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

 

Constraints:

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums is a non-decreasing array.
• -109 <= target <= 109

Related Topics

数组

二分查找

👍 3021

👎 0

解法

时间复杂度为 O (log N) 并且是一个非降序数组，暗示需要采用二分查询方法

采用二分法找到 target 元素后，通过 while 循环进行命中的元素的边界扩展

# leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        left = 0
        right = len(nums)

        while left < right:
            mid = (left + right) // 2

            if nums[mid] == target:
                mid_l = mid_r = mid
                while mid_l > 0 and nums[mid_l - 1] == target: mid_l -= 1
                while mid_r < len(nums) - 1 and nums[mid_r + 1] == target: mid_r += 1

                return [mid_l, mid_r]

            elif nums[mid] > target:
                right = mid
            else:
                left = mid + 1

        return [-1, -1]

        
# leetcode submit region end(Prohibit modification and deletion)

复杂度分析

• 时间复杂度O(logN)
• 空间复杂度O(1)