34. 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,

return [3, 4].

【思路】STL上已经为我们实现好了 二分查找 函数的接口。对于“vector”容器，我们可以利用“lower_bound” 查找 目标数值 target 第一次出现的位置。如果原数组中并不存在这个 target，lower_bound会返回第一个比target值大的位置。利用lower_bound函数获取 target 在原数组的位置，而类似的函数upper_bound则会返回第一个比target大的位置。

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {       vector<int> re;       if(nums.size()==0) return re;       vector<int>::iterator it1 = lower_bound(nums.begin(), nums.end(), target);       vector<int>::iterator it2 = upper_bound(nums.begin(), nums.end(), target);              if(*it1!=target)       {           re.push_back(-1);           re.push_back(-1);           return re;       }else       {           re.push_back(it1 - nums.begin());           re.push_back(it2 - nums.begin() - 1);       }        return re;    }};

二分查找+递归

<pre name="code" class="cpp">class Solution {public:    int binarySearch(vector<int>nums, int first, int last, int target)    {                while(first<= last)        {            int mid = first + (last - first) /2;            if(nums[mid]==target)            {                return mid;            }else if(nums[mid]>target)              last = mid-1;             else              first = mid +1;        }        return -1;    }    vector<int> searchRange(vector<int>& nums, int target) {       /*vector<int> re;       if(nums.size()==0) return re;       vector<int>::iterator it1 = lower_bound(nums.begin(), nums.end(), target);       vector<int>::iterator it2 = upper_bound(nums.begin(), nums.end(), target);              if(*it1!=target)       {           re.push_back(-1);           re.push_back(-1);           return re;       }else       {           re.push_back(it1 - nums.begin());           re.push_back(it2 - nums.begin() - 1);       }        return re;*/    //二分查找        vector<int> ret(2,-1);       if(nums.size()==0) return ret;       int low = 0, high = nums.size()-1;       int left;       int right;        while((left = binarySearch(nums, low, high, target)) != -1)        {            ret[0] = left;            high = left-1;        }        low = 0, high = nums.size()-1;        while((right = binarySearch(nums, low, high, target)) != -1)        {            ret[1] = right;            low = right+1;        }        return ret;    }};