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

算法：

1）使用Hash，存储各个值出现的下表，O(N)；

2）使用排序，O(NlogN）。

c++，O(N):

class Solution {
public:
    
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int> v;
        map<int,int> m;
        for(int i=0;i<numbers.size();i++){
            if(m.find(target-numbers[i])!=m.end()){
                int i1=i+1;//
                int i2=m.find(target-numbers[i])->second;
                v.push_back(min(i1,i2));
                v.push_back(max(i1,i2));
                return v;
            }
            m[numbers[i]]=i+1;//序号从1开始
        }
    }
};

java: O(NlogN)

public class Solution {
    class Node{
      int pos;
      int val;
    };
    public int[] twoSum(int[] numbers, int target) {
        Node[] nodes = new Node[numbers.length];
        for(int i=0;i<numbers.length;i++){
            nodes[i] = new Node();
            nodes[i].pos=i+1;
            nodes[i].val=numbers[i];
        }
        Arrays.sort(nodes, new Comparator<Node>(){
            public int compare(Node n1,Node n2){
                return n1.val-n2.val;
            }
        });
        int[] r = new int[2];
        int i=0;
        int j=nodes.length-1;
        while(i<j){
            int sum =nodes[i].val+nodes[j].val;
            if(sum==target){
                int max=Math.max(nodes[i].pos,nodes[j].pos);
                int min=Math.min(nodes[i].pos,nodes[j].pos);
                r[0]=min;
                r[1]=max;
                break;
            }else if(sum<target){
                i++;
            }else{
                j--;
            }
        }
        return r;
    }
}

Two Sum