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.
Note:
Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
：：
给定一个有序数组和一个目标值
找出数组中两个成员，两者之和为目标值，并顺序输出
假设一定存在一个解
//二分法查找实现
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target)
{
vector<int> res;
int n = numbers.size();
for (int i = 0; i < n; i++)
{
int left = i , right = n - 1;
while (left <= right)
{
int mid = (left + right) / 2;
if (numbers[mid]+ numbers[i] < target)
left = mid + 1;
else if (numbers[mid]+ numbers[i] > target )
right = mid - 1;
else //否则相等
{
res.push_back(i+1); //加1的原因题目输出要求决定的
res.push_back(mid+1);
return res;
}
}
}
return res;
}
};
//双指针实现
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target)
{
vector<int> res;
int left = 0, right = numbers.size() - 1;//定义两个指示变量
while (left < right) {
if (numbers[left] + numbers[right] < target)
left++;
else if (numbers[left] + numbers[right] > target)
right--;
else
{
res.push_back(left + 1), res.push_back(right + 1);
return res;
};
}
return res;
}
};
