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].

 

解法：find函数，返回找到数字的迭代器，然后用distance函数返回下标值，后面就是循环找出相同的个数

class Solution {public:    vector
     
       searchRange(vector
      
       & nums, int target) {        vector
       
        ::iterator it;        vector
        
          result;        it = find(nums.begin(),nums.end(),target);        if(it == nums.end()) {            result.push_back(-1);            result.push_back(-1);            return result;        }                int i = distance(nums.begin(),it);        int n = nums.size();        result.push_back(i);        while(*it == nums[++i] && i < n);  //确保++i不要数组溢出        result.push_back(i-1);        return result;    }};