leetcode No1. Two Sum

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Question：

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

即，给一个整型数组nums和一个整数target，找出两个数相加等于target，并返回这两个数在nums的下标（假定这两个数一定存在）

Algorithm：Hash Table

1、用Hash Table统计每个元素出现的次数

2、设x+y = target，第一次循环，找出x，即hash[x]>0&&hash[y]>0

3、第二次循环从x的下标开始，找y的下标

Submitted Code：

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int,int> hash;        vector<int> twoSum;        for(int &i : nums)            hash[i]++;        int m=0,y=0;        for(int i=0;i<nums.size();i++)        {            int x = nums[i];            y = target - x;            if(hash[y] > 0 && x!=y)            {                m = i;                twoSum.push_back(i);                break;            }            if(hash[y] > 1 && x==y)   //x==y的情况            {                m = i;                twoSum.push_back(i);                break;            }        }        for(int j=m+1;j<nums.size();j++)        {            if(nums[j] == y)            {                twoSum.push_back(j);                break;            }        }        return twoSum;    }};

2017/3/2更新

可以一遍遍历就得出结果，还是用hashtable，边遍历，边记录

因为X+Y=target，第一个数X可以记录在哈希表中，等到遍历到Y时，哈希表中已经有X了

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int,int> hash;   //Key is the number and value is its index in the vector        vector<int> res;        int numToFind=0;        for(int i=0;i<nums.size();i++)        {            numToFind=target-nums[i];            if(hash.find(numToFind)!=hash.end()) //if numberToFind is found in map, return them            {                res.push_back(hash[numToFind]);                res.push_back(i);                return res;            }            //number was not found. Put it in the map.            hash[nums[i]]=i;        }        return res;    }};

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