[leetcode]Two Sum

[leetcode]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：

    先将vector排序，然后分别从头尾元素开始，不停向中间逼近，直到有两个值相加等于target为止。

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target)     {        vector<int> result;        if (nums.size() < 2)            return result;        vector<int> temp(nums);        sort(nums.begin(), nums.end());        auto index1 = 0;        auto index2 = nums.size()-1;        while (index1 < index2)        {            if (nums[index1] + nums[index2]>target)            {                --index2;            }            else if (nums[index1] + nums[index2] < target)            {                ++index1;            }            else            {                for (auto i = 0; i < temp.size(); i++)                {                    if (nums[index1] == temp[i])                    {                        result.push_back(i + 1);                        break;                    }                }                for (auto i = temp.size() - 1; i >= 0; i--)                {                    if (nums[index2] == temp[i])                    {                        result.push_back(i + 1);                        break;                    }                }                break;            }        }        if (result[0] > result[1])            swap(result[0],result[1]);        return result;    }};

不足：用了一个跟nums一模一样的辅助数组来记录元素的原始位置。

另一种解法

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int, int> m;        for(int i = 0; i < nums.size(); i++)        {            if(m.find(target-nums[i]) != m.end())            return {m[target-nums[i]], i};            m[nums[i]] = i;        }    }};

不过map里会产生一些无用的数据。