LeetCode #34 - Search for a Range

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]

.
有序数组的查找，使用二分法，只是需要求查找值的所有元素的范围，所以可以先查找出一个元素，然后向两边遍历，确定所有元素的范围。
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int begin=0;
int end=nums.size()-1;
int mid=(begin+end)/2;
while(begin<=end)
{
if(nums[mid]==target)
{
int i=mid;
int j=mid;
while(nums[i-1]==target&&i>0) i--;
while(nums[j+1]==target&&j<(nums.size()-1)) j++;
vector<int> result;
result.push_back(i);
result.push_back(j);
return result;
}
else if(target>nums[mid])
{
begin=mid+1;
mid=(begin+end)/2;
}
else if(target<nums[mid])
{
end=mid-1;
mid=(begin+end)/2;
}
}
return vector<int>(2,-1);
}
};