34. Search for a Range（二分查找有重复元素数组中的目标数的第一个位置和最后一个位置）

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]

.

1.我的解答 二分查找

class Solution {
public:
int searchfirst(int num, vector<int>& vec){
int begin = 0;
int end = vec.size()-1;
int middle;
while(begin < end - 1){
middle = begin+(end-begin)/2;
if(vec[middle] >= num){
end = middle;
}else
begin = middle;
}
if(vec[begin] == num) return begin;
else return end;
}

int searchback(int num, vector<int>& vec){
int begin = 0;
int end = vec.size()-1;
int middle;
while(begin < end - 1){
middle = begin+(end-begin)/2;
if(vec[middle] > num){
end = middle;
}else
begin = middle;
}
if(vec[end] == num) return end;
else return begin;
}

vector<int> searchRange(vector<int>& nums, int target) {
vector<int>res;
int firstpos = searchfirst(target, nums);
if(nums[firstpos] != target)
res.push_back(-1);
else
res.push_back(firstpos);
int backpos = searchback(target, nums);
if(nums[backpos] != target)
res.push_back(-1);
else
res.push_back(backpos);
return res;
}
};