Problem:

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

Summary:

有序数组,找到两数之和等于target的数字下标。

Solution:

两个指针i和j分别指向数组头和尾,若nums[i] + nums[j] > target,则j--,反之则i++,直至找到目标数。

 class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int len = numbers.size();
int i = , j = len - ;
vector<int> res;
while (i < j) {
if (numbers[i] + numbers[j] > target) {
j--;
}
else if (numbers[i] + numbers[j] < target) {
i++;
}
else {
return {i + , j + };
}
} return {};
}
};

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