Two Sum

题目描述：

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

相信当你看见这篇博客时，已经独立思考过，这里直接给出Accepted Code。部分代码来自网络，希望不会涉及版权问题。

Solution1:

vector<int> twoSum(vector<int>& nums, int target) {
    vector<int> res(2);
    int n = nums.size();
    if(n < 2)
        return res;
    unordered_map<int, int> map;
    unordered_map<int, int>::const_iterator it = map.end();
    int tmp;
    for (int i = 0;i < n;++i)
    {
        tmp = target - nums[i];
        it = map.find(tmp);
        if (it != map.end())
        {
            res[0] = it->second + 1;
            res[1] = i + 1;
            return res;
        }
        else
            map[nums[i]] = i;
    }
}

 

后记：

　　这道题还有一种解法就是先排序，然后首尾两个指针遍历寻找。上述使用hash表的解法，其原理依旧是暴力破解，并不算好。(今天做3Sum有感)

solution2：

struct Node{
    int id,val;
};
bool compare(const Node &a, const Node &b){
    return a.val < b.val;
}

vector<int> twoSum(vector<int> &numbers, int target) {
    vector<Node> nodes(numbers.size());
    for(int i = 0;i < numbers.size();++i){
        nodes[i].id = i+1;
        nodes[i].val = numbers[i];
    }
    sort(nodes.begin(),nodes.end(),compare);
    int start = 0, end = numbers.size()-1;
    vector<int> res;
    while(start < end){
        if(nodes[start].val + nodes[end].val == target){
            if(nodes[start].id > nodes[end].id)
                swap(nodes[start].id , nodes[end].id);
            res.push_back(nodes[start].id);
            res.push_back(nodes[end].id);
            return res;
        }
        else if( nodes[start].val + nodes[end].val < target )
            ++start;
        else
            --end;
    }
}