Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
从i开始枚举，然后从i+1至.size()-1二分查找，好笨的方法=。=
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
vector<int>vec;
for (vector<int>::size_type i = 0; i < numbers.size(); ++i)
{
left = i + 1;
right = numbers.size() - 1;
int result = BinarySearch(numbers, target - numbers[i]);
if (numbers[i] + numbers[result] == target)
{
//cout << numbers[i] << " " << numbers[result] << endl;
vec.push_back(i + 1);
vec.push_back(result + 1);
break;
}
}
return vec;
}
private:
int left;
int right;
int mid;
#define max(a,b) (((a) > (b)) ? (a) : (b))
#define min(a,b) (((a) > (b)) ? (b) : (a))
int BinarySearch(vector<int>& numbers, int target)
{
while(left <= right)
{
mid = (left + right) / 2;
if(numbers[mid] > target)
{
right = mid - 1;
}
else if(numbers[mid] < target)
{
left = mid + 1;
}
else
{
return mid;
}
}
return right;
}
};
int main()
{
Solution s;
vector<int>vec{1,2,3,4,4,9,56,90};
vector<int>v = s.twoSum(vec, 8);
for (vector<int>::size_type i = 0; i < v.size(); ++i)
{
cout << v[i] << endl;
}
return 0;
}
15 / 15 test cases passed.
Status:
Accepted
Runtime: 13 ms
Submitted: 2 minutes ago