Two Sum

Difficulty:Easy

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].

MySolution: O(n²)

vector<int> twoSum(vector<int>& nums, int target) {        vector<int> a(2);    //※函数返回类型是vector<int>，所以需要定义此类型变量         //nums.size()和sizeof(nums)返回的值是不同的        for(int i = 0; i<nums.size(); i++)           {            for(int j = 0; j<nums.size(); j++)            {                if((i != j) && (nums[i]+nums[j] == target))                {                    a[0] = i;                    a[1] = j;                          break;                }            }        }         return a;    }

OtherSolutions:O(n)

vector<int> twoSum(vector<int> &numbers, int target){    //Key is the number and value is its index in the vector.unordered_map<int, int> hash;vector<int> result;for (int i = 0; i < numbers.size(); i++) {int numberToFind = target - numbers[i];//遍历vector，计算出每个值跟目标值的差值，再用hash函数查找差值            //if numberToFind is found in map, return themif (hash.find(numberToFind) != hash.end()) {result.push_back(hash[numberToFind]);//将符合条件的数据在vector中的位置存到result中并返回resultresult.push_back(i);return result;}            //number was not found. Put it in the map.hash[numbers[i]] = i;//将vector中的值作为map的键，对应的index作为map的值，在循环中，比较并查找符合条件的值得同时，在map中添加键值对,这种方法需要两个值都在map中}return result;}

public class Solution {    public int[] twoSum(int[] numbers, int target) {                HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>();        for(int i = 0; i < numbers.length; i++){            Integer diff = (Integer)(target - numbers[i]);            if(hash.containsKey(diff)){                int toReturn[] = {hash.get(diff)+1, i+1}; //在LeetCode提交时候不+1才能通过                return toReturn;            }            hash.put(numbers[i], i);        }                return null;            }}

class Solution(object):    def twoSum(self, nums, target):        if len(nums) <= 1:            return False        buff_dict = {}        for i in range(len(nums)):            if nums[i] in buff_dict:                return [buff_dict[nums[i]], i]//buff_dict[nums[i]]返回的是target - nums[i]的位置j，i为nums[i]的位置i            else:                buff_dict[target - nums[i]] = i//字典里存键值对,nums[i]与target的差值为字典的键，nums[i]在nums中的位置i为值,这种方法不用把nums[i]和对应的target-nums[i]全存进字典

Extended：

 HashMap实现原理分析

C++ unordered_map

map/unordered_map原理和使用整理