Search for a Range一打简单的二分题

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].

思路：这是一道简单的二分题，只要考虑好边界，二分即可。

代码如下：

vector<int> searchRange(vector<int>& nums, int target) {
int begin=0;
vector<int>temp;
if(nums.size()==0){
temp.push_back(-1);
temp.push_back(-1);
return temp;
}
int end=nums.size()-1;
int mid=(end+begin)/2;
while(begin<end){
if(nums[mid]==target) break;
if(nums[mid]<target){
if(mid==begin){
mid=end;
break;
}
begin=mid;
mid=(end+begin)/2;
}
if(nums[mid]>target){
if(mid==end) {
mid=begin;
break;
}
end=mid;
mid=(end+begin)/2;
}
}
int a,b;
if(nums[mid]==target){
a=mid;
b=mid;
while(nums[a]==target&&a>=0){
a--;
}
while(nums[b]==target){
b++;
}

temp.push_back(a+1);
temp.push_back(b-1);
}
else{

temp.push_back(-1);
temp.push_back(-1);
return temp;

}
return temp;
}
};