[LeetCode] Two Sum

问题：找出数组中两个数之和等于一个给定目标数。并输出这两个数在数组中的下标。

1.暴力O(n^2)，两个for循环查找，TLE。

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
  
        vector<int> ans;
        for (int i = 0; i < numbers.size()-1; i++)
        {
            for (int j = i+1; j < numbers.size(); j++)
            {
                if (numbers[i] + numbers[j] == target)
                {
                    ans.push_back(i+1);
                    ans.push_back(j+1);
                    return ans;
                }
            }
        }
    }
};

2.O(n*logn),用一个pair，记录index和num，然后排序num，用两个指针向中间移动寻找答案

如果 sum = l.num + r.num, 那么 l.index 和 r.index 就是答案。注意答案index的顺序。

int cmp(const pair<int, int> &a, const pair<int, int> &b) {
     if (a.first == b.first)     return a.second < b.second;
     else return a.first < b.first;
}
 
class Solution {
public:
 
     vector<int> twoSum(vector<int> &numbers, int target) {
     
          assert(numbers.size() >= 2);
 
          vector<int> ans;
          vector< pair<int, int> > PairNumber;
          for (int i = 0; i < numbers.size(); i++) {
               PairNumber.push_back(make_pair(numbers[i], i+1));
          }
 
          sort(PairNumber.begin(), PairNumber.end(), cmp);
 
          int l = 0, r = PairNumber.size()-1;
          while (l < r) {
               int sum = PairNumber[l].first + PairNumber[r].first;
               if (sum == target) {
                    ans.push_back(min(PairNumber[l].second, PairNumber[r].second));
                    ans.push_back(max(PairNumber[l].second, PairNumber[r].second));
                    break;
               }
               else if (sum < target) {
                    l++;
               }
               else {
                    r--;
               }
          }
          return ans;
     }
};

3.O(n)，用hash来做。C++中使用map,O（nlogn）。在边进行hash当前数numbers[i]的时候边判断是否出现target-numbers[i]。

C++:

class Solution {
public:
 
     vector<int> twoSum(vector<int> &numbers, int target) {
     
          vector<int> ans;
          map<int, int> hash;
          for (int i = 0; i < numbers.size(); ++i) {
               if (hash.find(target-numbers[i]) != hash.end()) {
                    ans.push_back(hash[target-numbers[i]]);
                    ans.push_back(i+1);
                    break;
               }
               else {
                    hash.insert(make_pair(numbers[i], i+1));
               }
               
          }
          return ans;
     }
};

Java:

import java.util.Hashtable;
public class Solution {
     public int[] twoSum(int[] numbers, int target) {
          assert (numbers.length >= 2);
          int ans[] = new int[2];
          Hashtable<Integer, Integer> hash= new Hashtable<Integer, Integer>();
          for (int i = 0; i < numbers.length; i++) {
               if (hash.get(target-numbers[i]) != null) {
                    ans[0] = hash.get(target-numbers[i]);
                    ans[1] = i+1;
                    break;
               }
               else {
                    hash.put(numbers[i], i+1);
               }
          }
          return ans;
     }
}