LeetCode twoSum

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.

 

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大致的问题：给定一个整形数组，返回一个两数之和等于给定目标的目录。你可以假定每次输入只有一个解决办法

 

我首先想到的是，一个一个的解决，时间复杂度为（n²),空间复杂度为（1）

代码如下，结果是当数组的长度过大时，会超时。

class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
    vector<int> results;
    auto len = nums.size();
    for (int i = 0; i<len || nums[i] >= target; ++i)
    {
        int j = i + 1;
        for (; j<len || nums[j] > target; ++j)
        {
            if ((nums[i] + nums[j]) == target)
            {
                results.push_back(i);
                results.push_back(j);
                break;
            }
        }
        if ((nums[i] + nums[j]) == target)
            break;
    }
    return results;
}
}

只能牺牲空间复杂度，来提升时间复杂度。而且题目中的tags有hash table，在C++中可以用map来代替，于是求得下列结果

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

        map<int, int>::iterator it;
        for (int i = 0; i < nums.size(); i++) {
            if ((it = m.find(target - nums[i])) != m.end())
            {
                if (i == it->second) continue;
                result.push_back(i);
                result.push_back(it->second);//将结果放入到result中
                return result;
            }
        }
        return result;
    }
};