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]`.

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``` //题目要求：给定一个已经排序的数组，求出target第一次出现的位置和最后出现的位置
//二分法思想，分别找到target第一次出现的位置index1，和target最后出现的位置index2
//利用(target+1)第一次出现的位置减1，得到target最后出现的位置
class Solution {
public:
vector searchRange(vector& nums, int target) {
int index1 = lower_bound(nums, target);
int index2 = lower_bound(nums, target + 1) - 1; //利用(target+1)第一次出现的位置减1，得到target最后出现的位置
if (index1 < nums.size() && nums[index1] == target)
return{ index1, index2 };
else
return{ -1, -1 };
}

int lower_bound(vector& nums, int target) {
int left = 0, right = nums.size() - 1;
while (left <= right)
{
int mid = (right + left) / 2;
if (nums[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
return left;
}
};```