[Lintcode]61. Search for a Range/[Leetcode]34. Find First and Last Position of Element in Sorted Array

• 本题难度: Medium/Medium

• Topic: Binary Search

  Description

  Given a sorted array of n integers,find the starting and ending position of a given target value.

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

Example
Example 1:

Input:
[]
9
Output:
[-1,-1]

Example 2:

Input:
[5,7,8,10]
8
Output:
[3,4]
Challenge
O(log n) time.

我的代码

class Solution:
    def searchRange(self,nums,target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        l = 0
        r = len(nums) - 1
        m = int((l + r) / 2)
        r1 = -1
        r2 = -1
        while (l <= r):
            if nums[m] == target and (m == 0 or nums[m - 1] < target):
                    r1 = m
                    r2 = m
                    l += 1
                    break
            else:
                if nums[m] >= target:
                    r = m - 1
                else:
                    l = m + 1
                m = int((l + r) / 2)
        r = len(nums) - 1
        while (l <= r):
            if nums[m] == target and (m == r or nums[m + 1] > target):
                r2 = m
                break
            else:
                if nums[m] <= target:
                    l = m + 1
                else:
                    r = m - 1
                m = int((l + r) / 2)
        return [r1,r2]

别人代码

def searchRange(self,target):
    def search(n):
        lo,hi = 0,len(nums)
        while lo < hi:
            mid = (lo + hi) / 2
            if nums[mid] >= n:
                hi = mid
            else:
                lo = mid + 1
        return lo
    lo = search(target)
    return [lo,search(target+1)-1] if target in nums[lo:lo+1] else [-1,-1]# 处理得很舒服

def searchRange(self,target):
    lo = bisect.bisect_left(nums,target)
    return [lo,bisect.bisect(nums,target)-1] if target in nums[lo:lo+1] else [-1,-1]

思路

同类问题[Lintcode] 14. First Position of Target 但要注意，最后一个元素和第一个元素一个二分查找法是无法实现的。

• 时间复杂度: O(log(n))