LeetCode#1.Two Sum

题目

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

思路

第一个想法是找出在给出的数中两两组合，若和为目标则存入结果。但这样复杂度为O(n2)，显然需要改进。

改进的思路：使用哈希函数的思想。用map存储对，扫描数组，对于当前的数算出另一个数，如果map中有这个记录，则返回结果。否则把当前的数存入map，置其值为序号。

代码

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        if (nums.size() == 0)
            return *new vector<int>(0, 0);
        map<int, size_t> hash;
        for (int i = 0; i < nums.size(); ++i) {
            int other = target - nums[i];
            if (hash[other] != 0) {
                vector<int> result;
                result.push_back(hash[target - nums[i]]-1);
                result.push_back(i);
                return result;
            }
            else
                hash[nums[i]] = i+1;
        }
        vector<int> tmp;
        return tmp;
    }