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

二分

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

分析

二分找边界

bisect target不存在时 left=right

from bisect import bisect_left, bisect_right


class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        left = bisect.bisect_left(nums, target)
        right = bisect.bisect_right(nums, target)
        if left == right:
            return [-1, -1]
        else:
            return [left, right-1]

九章模板必须分别判断s,e

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        if not nums:
            return [-1, -1]
        s,e = 0,len(nums)-1
        while s + 1 < e:
            mid = (s + e) // 2
            if nums[mid] >= target:
                e = mid
            else:
                s = mid
        if nums[s] == target:
            start = s
        elif nums[e] == target:
            start = e
        else:
            return [-1,-1]

        s, e = 0, len(nums) - 1
        while s + 1 < e:
            mid = (s + e) // 2
            if nums[mid] <= target:
                s = mid
            else:
                e = mid
        if nums[e] == target:
            end = e
        else:
            end = s
        
        return [start,end]