1. Two Sum

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.

public class Solution {
public int[] TwoSum(int[] nums, int target) {
int[] sumnumbers =new int[]; for (int i = ; i <= nums.Length-; i++)
{
for (int j = ; j <= nums.Length-; j++)
{
if (nums[i] + nums[j] - target == && i != j)
{
if (i < j)
{
sumnumbers[] = i;
sumnumbers[] = j;
return sumnumbers;
}
else
{
sumnumbers[] =j;
sumnumbers[] = i;
return sumnumbers;
}
} }
}
return sumnumbers;
}
}

Top sulution 【O(n)】 C++的top解决方案

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 = ; i < numbers.size(); i++) {
int numberToFind = target - numbers[i]; //if numberToFind is found in map, return them
if (hash.find(numberToFind) != hash.end()) {
//+1 because indices are NOT zero based
result.push_back(hash[numberToFind] + );
result.push_back(i + );
return result;
} //number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
}

C#的仿写

public class Solution {
public int[] TwoSum(int[] nums, int target) {
int[] RetIndecis = {,};
Hashtable hsNums = new Hashtable();
hsNums.Clear();
//i is the latter index
for(int i=;i<nums.Length;i++)
{
int targetKey = target - nums[i];
//if targetKey exist
if(hsNums.ContainsKey(targetKey))
{
RetIndecis[] = i + ;
RetIndecis[] = (int)hsNums[targetKey];
return RetIndecis;
}
//key is the number and value is the index,filter the number which has existed
if(!hsNums.ContainsKey(nums[i]))
hsNums.Add(nums[i],i + );
}
return(RetIndecis);
}
}

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