[LeetCode]题解（python）：034-Search for a Range

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题目来源

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https://leetcode.com/problems/search-for-a-range/

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

Your algorithm's runtime complexity must be in the order of O(log n).

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

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题意分析

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Input: a list and a target(int)

Output: a list with the first index and the last index of target in the input list

Conditions:时间复杂度为0（logn），两个index分别为起始和最后位置

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题目思路

本题可采用二分查找的算法，复杂度是0（logn）,首先通过二分查找找到taregt出现的某一个位置，然后以这个位置，以及此时的（first，last）来二分查找最左出现的位置和最右出现的位置

PS：注意边界条件

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AC代码（Python）

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 _author_ = "YE"
 # -*- coding:utf-8 -*-
 class Solution(object):
     def findRight(self, nums, first, mid):
         if nums[first] == nums[mid]:
             return mid
         nmid = (first + mid) // 2
         if nmid == first:
             return first
         if nums[nmid] == nums[first]:
             return self.findRight(nums, nmid, mid)
         else:
             return self.findRight(nums,first, nmid)

     def findLeft(self, nums, mid, last):
         if nums[mid] == nums[last]:
             return mid
         nmid = (mid + last) // 2
         if nmid == mid:
             return last
         if nums[nmid] == nums[last]:
             return self.findLeft(nums, mid, nmid)
         else:
             return self.findLeft(nums,nmid + 1, last)

     def searchRange(self, nums, target):
         """
         :type nums: List[int]
         :type target: int
         :rtype: List[int]
         """
         last = len(nums)
         first = 0
         while first < last:
             mid = (first + last) // 2
             # print(first,mid,last)
             if nums[mid] == target:
                 # print(self.findLeft(nums,first, mid))
                 # print(self.findRight(nums,mid,last - 1))
                 return [self.findLeft(nums,first, mid), self.findRight(nums,mid,last - 1)]
             elif nums[mid] < target:
                 first = mid + 1
             else:
                 last = mid

         return [-1,-1]