问题描述：
Given an array of integers, 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
解决原理：
数组以升序排序
从数组S头、尾同时遍历数组，i为头端索引，j为尾端索引
若S[i]+S[j]=target，则返回对应索引
若S[i]+S[j]>target，则j--
否则i++
代码：
1 class Solution { 2 public: 3 vector<int> twoSum(vector<int> &numbers, int target) { 4 int i = 0; 5 int j = numbers.size()-1; 6 int sum; 7 vector<int> rs; 8 vector<int> tmp; 9 10 for(int k = 0; k < numbers.size(); k++) 11 tmp.push_back(numbers[k]); 12 13 sort(tmp.begin(), tmp.end()); 14 while(i < j){ 15 sum = tmp[i] + tmp[j]; 16 if(sum == target){ 17 int m = 0; 18 while(numbers[m] != tmp[i]) m++; 19 int n = 0; 20 while(numbers[n] != tmp[j]) n++; 21 if(n == m){ 22 n++; 23 while(numbers[n] != tmp[j]) n++; 24 } 25 26 rs.push_back(m>n?n+1:m+1); 27 rs.push_back(m>n?m+1:n+1); 28 return rs; 29 }else if(sum < target){ 30 i++; 31 }else{ 32 j--; 33 } 34 } 35 return rs; 36 } 37 };