#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].

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<int> searchRange(vector<int>& nums, int target) {
        vector<int>::iterator it;
        vector<int> 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;
    }
};