LeetCode Search for a Range

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]

.

思路：进行二分搜索，当找到目标值时，在已该元素为中心进行扩展，从而确定区域范围

代码如下：class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int middle,front,back;
int start=-1,end=-1;
vector<int> result;
front = 0,back=nums.size()-1;
middle = (front+back)/2;
while(front<=back)
{
if(nums[middle] > target)
{
back=middle-1;
middle=(front+back)/2;
}
else if(nums[middle] < target)
{
front = middle+1;
middle=(front+back)/2;
}
else
{
start=end=middle;
while(start>0 && nums[start-1] == target)
start--;
while(end < nums.size()-1 && nums[end+1] == target)

4000
end++;
break;
}
}
result.push_back(start);
result.push_back(end);
return result;
}
};