【LeetCode 1_数组_哈希表】Two Sum

解法一：O(N)

 vector<int> twoSum(vector<int>& nums, int target)
 {
     unordered_map<int, int> numsHash;
     vector<int> res;

     for (int i = ; i < nums.size(); ++i) {
         int num_last = target - nums[i];
         if (numsHash.find(num_last) != numsHash.end()) {
             res.push_back(numsHash[num_last] + );
             res.push_back(i + );
             break;
         } else {
             numsHash[nums[i]] = i;
         }
     }
     return res;
 }

解法二：O(NlogN)

 vector<int> twoSum(vector<int>& nums, int target)
 {
     vector<int> res;
     int lpos = , rpos = nums.size() - ;
     sort(nums.begin(), nums.end());
     while (lpos < rpos) {
         int num = nums[lpos] + nums[rpos];
         if (num == target) {
             res.push_back(lpos + );
             res.push_back(rpos + );
         } else if (num > target) {
             rpos--;
         } else {
             lpos++;
         }
     }
     return res;
 }

解法二也是不错的思路，空间复杂度O(1)，但这个解法会超时。