LeetCode Two Sum

自己写的，运行28ms

vector<int> twoSum(vector<int> &numbers, int target) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.int i,j;vector<int> ret;for (i=0;i<numbers.size();i++){for (j=i+1;j<numbers.size();j++){if (numbers[i]+numbers[j]==target){ret.push_back(i);ret.push_back(j);return ret;}}}return ret;}

别人的，先排序，运行16ms，内部排序时间应该是O(NlogN)，http://blog.csdn.net/ju136/article/details/6975666

class node  {  public:      int id;      int value;  public:      node(int a, int b):id(a), value(b){}      friend bool operator < (const node &a, const node &b) { return a.value < b.value;}  };    vector<int> twoSum(vector<int> vs, int target)  {      const int s = vs.size();      vector<int> ret;      vector<node> temp;      for (int i = 0; i < s; ++i)          temp.push_back(node(i+1, vs[i]));      sort(temp.begin(), temp.end());        for (int b = 0, e = s - 1; b < e; )      {          const int v = temp[b].value + temp[e].value;          if (v == target)          {              if (temp[b].id < temp[e].id)              {                  ret.push_back(temp[b].id);                  ret.push_back(temp[e].id);              }              else              {                  ret.push_back(temp[e].id);                  ret.push_back(temp[b].id);              }                            return ret;          }          else if (v < target)          {              ++b;          }          else --e;      }      return ret;  }  

别人的，用map，时间也是16ms，因为map是用红黑树实现的，所以快在find算法了。http://blog.csdn.net/doc_sgl/article/details/12115067

vector<int> twoSum(vector<int> &numbers, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<int> res;    int length = numbers.size();if(length < 2)return res;map<int,int> mp;for(int i = 0; i < length; ++i)mp[numbers[i]] = i;map<int,int>::iterator it = mp.end();for(int i = 0; i < length; ++i){it = mp.find(target - numbers[i]);if(it != mp.end()){res.push_back(min(i+1,it->second +1));res.push_back(max(i+1,it->second +1));break;}}return res;    }

题目说第一个索引要小于第二个（应该是小于等于），其实，这题当输入只有一个时候同样元素时，比如4，2，5，找8，算会输出同样的1,1。