【LeetCode】12.Array and String — Input array is sorted 查找中枢索引

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.

根据体育首先想到了一个二层循环，结合章节要求的two-pointer,所以定义两个指针用来遍历数组。最主要的问题在于循环条件的设置。具体内容见代码及相关注释，不再详述。

 1 class Solution {
 2 public:
 3 vector<int> twoSum(vector<int>&numbers, int target) {
 4     int *index1 = &numbers[0];//定义两个指针,指向同一个数组
 5     int *index2 = &numbers[0];
 6     int counter1 = 0;//用来记录数组下标
 7     while (2*index1[counter1]<=target) {//以index1为基准，未达到括号里的条件则无需继续循环
 8         for (int i = counter1+1; i < numbers.size(); i++)//从index1往后进行遍历寻找
 9         {
10             if (index1[counter1] + index2[i] == target) {//找到则返回条件值，并处理指针；
11                 index1 = NULL;
12                 index2 = NULL;
13                 return{ counter1+1,i+1 };
14             }
15             else if (index1[counter1] + index2[i] > target) break; //否则向后移动继续寻找
16         }
17         counter1++;
18     }
19     index1 = NULL;
20     index2 = NULL;
21     return{};//未找到，返回空。
22 }
23 };