[leetcode]34. Search for a Range

题目链接：https://leetcode.com/problems/search-for-a-range/#/description

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]

.

方法一：

class Solution{
public:
vector<int> searchRange(vector<int>& nums,int target)
{
vector<int> res(2,-1);
if(nums.empty())
return res;
int i=0,j=nums.size()-1;
//找到左边起始位置
while(i<j)
{
int mid=i+(j-i)/2;
if(nums[mid]<target)
i=mid+1;
else
j=mid;
}
if(nums[i]!=target)
return res;
else
res[0]=i;
//找到右边结束位置
j=nums.size()-1;
while(i<j)
{
int mid=i+(j-i)/2+1;
if(nums[mid]>target)
j=mid-1;
else
i=mid;
}
res[1]=j;
return res;
}
};

方法二：
class Solution{
public:
vector<int> searchRange(vector<int>& nums,int target)
{
auto bounds=equal_range(nums.begin(),nums.end(),target);
if(bounds.first==bounds.second)
return {-1,-1};
return {bounds.first-nums.begin(),bounds.second-nums.begin()-1};
}
};