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

注意1.索引號從1開始

       2.只有一個解

解題思路

A方案（兩層迴圈）O(n2)

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        vector<int> result;
         for (int i = 0; i < numbers.size()-1; i++) {
             for (int j = i+1; j < numbers.size(); j++) {
                 if (numbers[i] + numbers[j]==target) {
                     result.push_back(i);
                     result.push_back(j);
                     return result;
                 }
             }
         }
         return result;
        
    }
};
 

B方案O(nlogn)。排序，然後兩個指標一前一後。因為題中說明了只有一對答案，因此不需要考慮重複的情況。

c方案 O(nlogn)用map結構來完成

map的相關介紹：http://blog.sina.com.cn/s/blog_61533c9b0100fa7w.html

class Solution {
public:
   vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int> result;
        map<int, int> m;
        if (numbers.size() < 2)
            return result;
        for (int i = 0; i < numbers.size(); i++)
            m[numbers[i]] = i;

        map<int, int>::iterator it;
        for (int i = 0; i < numbers.size(); i++) {
            if ((it = m.find(target - numbers[i])) != m.end())
            {
                if (i == it->second) continue;
                result.push_back(i);
                result.push_back(it->second);
                return result;
            }
        }
         return result;
        
    }
};