34. Search for a Range 解题思路

题目描述：

Given an array of integers sorted in ascending order, 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].

解题思路：

这题我用的方法比较笨：从两端搜索数组，两个都是target数才跳出。具体的就看代码吧。

代码：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int left = 0, right = nums.size()-1;
        vector<int> ret = {-1, -1};
        while(right >= left){
            if(nums[right] == nums[left] && nums[left] == target){
                ret[0] = left;
                ret[1] = right;
                break;
            }
            if(nums[left] != target)
                left++;
            if(nums[right] != target)
                right--;
        }
        return ret;
    }
};