LeetCode 1. Two Sum

https://leetcode.com/problems/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]. 

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• 数组简单题，hash表。有三种解法：1、跟冒泡思想一样，挨个做加法比较；2、先做hash，然后扫描一遍数组，用STL find看是否在hash表中能找到余数。这里需要注意如果全部扫描一遍，是可以找到两对重复结果的，所以如果找到结果就返回；3、跟2的思想一样用hash，不过针对2种会出现重复结果的情况，这次直接扫描数组，边扫描数组，边看是否hash表中能找到余数，如果没有则边插入到hash表。
• 注意对map插入值时，不能使用at(k)来访问下标插入，因为k这时候不存在，所以会报错。需要用[k]来代替。
• map::at - C++ Reference
  • http://www.cplusplus.com/reference/map/map/at/
  • Returns a reference to the mapped value of the element identified with key k. 
  • If k does not match the key of any element in the container, the function throws an out_of_range exception.
• https://leetcode.com/problems/two-sum/solution/

//
//  main.cpp
//  LeetCode
//
//  Created by Hao on 2017/3/16.
//  Copyright © 2017年 Hao. All rights reserved.
//

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

class Solution {
public:
    // Approach #1 (Brute Force)
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> vResult;
        
        for (int i = 0; i < nums.size(); i ++) {
            for (int j = i + 1; j < nums.size(); j ++) { // j = i + 1 to avoid dup
                if (target == nums.at(i) + nums.at(j)) {
                    vResult.push_back(i);
                    vResult.push_back(j);
                }
            }
        }
        
        return vResult;
    }

    // Approach #2 (Two-pass Hash Table)
    vector<int> twoSum2(vector<int>& nums, int target) {
        vector<int> vResult;
        unordered_map<int, int> hashmap;
        
        for (int i = 0; i < nums.size(); i ++) {
            // hashmap.at(nums.at(i)) = i; // terminating with uncaught exception of type std::out_of_range: unordered_map::at: key not found
            hashmap[nums.at(i)] = i;
        }
        
        for (int i = 0; i < nums.size(); i ++) {
            int complement = target - nums.at(i);
            
            if (hashmap.find(complement) != hashmap.end() && hashmap.at(complement) != i) {
                vResult.push_back(i);
                vResult.push_back(hashmap.at(complement));
                
                return vResult; // Need to return here, or else would find dup results (i, hashmap.at(complement)) & (hashmap.at(complement), i)
            }
        }
        
        return vResult;
    }

    // Approach #3 (One-pass Hash Table)
    vector<int> twoSum3(vector<int>& nums, int target) {
        vector<int> vResult;
        unordered_map<int, int> hashmap;
        
        for (int i = 0; i < nums.size(); i ++) {
            int complement = target - nums.at(i);
            
            if (hashmap.find(complement) != hashmap.end() && hashmap.at(complement) != i) {
                vResult.push_back(i);
                vResult.push_back(hashmap.at(complement));
            }
            
            hashmap[nums.at(i)] = i;
        }
        
        return vResult;
    }
};

int main(int argc, char* argv[])
{
    Solution    testSolution;
    string      result;
    
    vector<int> iVec = {2, 7, 11, 15};
    vector<int> vResult;

    /*
     0 1
     0 1
     1 0
     */
    vResult = testSolution.twoSum(iVec, 9);
    
    for (auto iter : vResult)
        cout << iter << " ";
    cout << endl;
    
    vResult.clear();
    
    vResult = testSolution.twoSum2(iVec, 9);
    
    for (auto iter : vResult)
        cout << iter << " ";
    cout << endl;

    vResult.clear();
    
    vResult = testSolution.twoSum3(iVec, 9);
    
    for (auto iter : vResult)
        cout << iter << " ";
    cout << endl;

    return 0;
}

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• Python3 Solution
• Built-in Types — Python 3.7.2 documentation - get(key[, default])
  • https://docs.python.org/3/library/stdtypes.html?highlight=get#dict.get
  • Return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError.
Built-in Functions — Python 3.7.2 documentation - enumerate(iterable, start=0)
• https://docs.python.org/3/library/functions.html?highlight=enumerate#enumerate
• Return an enumerate object. iterable must be a sequence, an iterator, or some other object which supports iteration. The __next__() method of the iterator returned by enumerate() returns a tuple containing a count (from start which defaults to 0) and the values obtained from iterating over iterable.

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        dict = {}
        
        for id, num in enumerate( nums ):            
            if dict.get( target - num ) is not None:
                return [ id, dict[ target - num ] ]
            else:            
                dict[ num ] = id