[Array]Two Sum

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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首先对数组进行排序，然后遍历数组，即针对每一个元素，寻找(target-nums[i]), 寻找的方法采用二分法。这样空间复杂度为O(n)(用来记录下标),排序时间复杂度为O(nlogn),遍历数组进行二分查找时间复杂度为O(nlogn)。

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> res;        vector<int> map(nums);        sort(nums.begin(),nums.end());        for(int i = 0;i < nums.size();i++){            int value = target - nums[i];            int left=i+1,right=nums.size()-1;            while(left<=right){                int middle = (left+right)/2;                if(nums[middle]==value){                    for(int j=0;j<map.size();j++){                        if(map[j]==nums[i]||map[j]==value)                            res.push_back(j);                    }                   break;                }                if(value<nums[middle]){                    right = middle-1;                }                else                    left = middle+1;             }            if(!res.empty())                break;        }        return res;    }};

另外一种想法是利用map，大致思路是遍历一遍数组，将已经访问过的元素存入到map中，再利用value = target - 正在访问的元素，然后回到map中进行查找value。

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> res;        map<int,int> mp;        map<int,int>::iterator it;        for(int i=0;i<nums.size();i++){            int find_number = target - nums[i];            it = mp.find(find_number);            if(it!=mp.end()){                res.push_back(it->second);                res.push_back(i);                break;            }            mp[nums[i]]=i;        }        return res;    }};