Leetcode: 1.Two Sum

Description

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, and you may not use the same element twice.

Example

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路

• 直接一個排序，O(nlgn)，然後一個查找，O(n)
• 使用unordered_map

代碼

• O(nlgn)的解法

  class Solution {
  public:
  vector<int> twoSum(vector<int>& nums, int target) {
      vector<int> res;
      int len = nums.size();
      if(len == 0) return res;
  
      vector<pair<int, int>> vec;
      for(int i = 0; i < len; ++i){
          vec.push_back(pair<int, int>(nums[i], i));
      }
  
      sort(vec.begin(), vec.end(), 
          [](const pair<int, int> &a, const pair<int, int>&b){ return a.first < b.first;});
  
      int i = 0, j = len - 1, tmp;
      while(i < j){
          tmp = vec[i].first + vec[j].first;
          if(tmp == target){
              res.push_back(min(vec[i].second, vec[j].second));
              res.push_back(max(vec[i].second, vec[j].second));
              break;
          }
          else if(tmp < target)
              i++;
          else j--;
      }
  
      return res;
  }
  };

• O(n)的解法

  class Solution {
  public:
  vector<int> twoSum(vector<int>& nums, int target) {
      vector<int> res;
      int len = nums.size();
      if(len == 0) return res;
  
      unordered_map<int, int> hash;
      int needToFind = 0;
      for(int i = 0; i < len; ++i){
          needToFind = target - nums[i];
          if(hash.find(needToFind) != hash.end()){
              res.push_back(min(i, hash[needToFind]));
              res.push_back(max(i, hash[needToFind]));
              break;
          }
          else{
              hash.insert(pair<int, int>(nums[i], i));
          }
      }
      return res;
  }
  };