zoukankan      html  css  js  c++  java
  • [LeetCode] Two Sum

    This is a classic problem for hash table. The basic idea is to maintain a hash table for each element in nums, using the element as key and its index (in this problem, 1-based) as value. Then for each num of nums, search for target - num in the hash table. If it is found and is not the same as num, then we are done. Notice that the problem statement has excluded the case of duplicates by stating that "each input would have exactly one solution".

    Now you may quickly write down the following code.

     1 vector<int> twoSum(vector<int>& nums, int target) {
     2     vector<int> ans;
     3     unordered_map<int, int> mp;
     4     for (int i = 0; i < nums.size(); i++)
     5         mp[nums[i]] = i + 1;
     6     for (int i = 0; i < nums.size(); i++) {
     7         if (mp.find(target - nums[i]) != mp.end() && mp[target - nums[i]] != i + 1) {
     8             ans.push_back(i + 1);
     9             ans.push_back(mp[target - nums[i]]);
    10             return ans;
    11         }
    12     }
    13 }

    Notice the above code has two for loops to iterate over nums. In fact, the process of building the hash table and searching in it can be done in one pass. Each time before you add a num to mp, just search for target - num first. The code now becomes as follows.

     1 vector<int> twoSum(vector<int>& nums, int target) {
     2     vector<int> ans;
     3     unordered_map<int, int> mp;
     4     for (int i = 0; i < nums.size(); i++) {
     5         if (mp.find(target - nums[i]) != mp.end() && mp[target - nums[i]] != i + 1) {
     6             ans.push_back(mp[target - nums[i]]);
     7             ans.push_back(i + 1);
     8             return ans;
     9         }
    10         mp[nums[i]] = i + 1; 
    11     }
    12 }
  • 相关阅读:
    【python】正则表达式
    Java 接口、抽象类
    设计模式之抽象工厂方法模式
    设计模式之工厂方法模式
    设计模式之单例模式
    pulltorefresh学习
    ProgressDialog使用
    android:descendantFocusability用法简析
    数据保存之File
    runOnUiThread学习
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4561484.html
Copyright © 2011-2022 走看看