LeetCode-Two Sum

LeetCode-Two Sum
Description:
Given an array of integers, find two numbers such that they add up to a specific target number.


The function twoSum shocould return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.


You may assume that each input wowould have exactly one solution.


Input: numbers = {2, 7, 11, 15}, target = 9
Output: index1 = 1, index2 = 2


A typical question is to find two numbers in an array so that their sum is the target.
There are several solutions to this question. The common solution is sorting + binary search. Here is a solution to use hash tables.
1. Save nums [I] and index [I] During array nums traversal. If nums [I] already exists:
1.1 If nums [I] = target/2, {hash [nums [I], I} is returned}
1.2 If not, ignore
2. traverse the hash key-value pairs and determine whether the hash [target-k] exists for each key k, if hash [target-k] exists and is not the same as hash [k] (not in the same position), the result is returned.
3. Pay attention to the front and back of small indexes when returning.


Implementation Code:

public class Solution {    public int[] TwoSum(int[] nums, int target)     {        var hash = new Dictionary
 
  ();    for(var i = 0;i < nums.Length; i++){    if(!hash.ContainsKey(nums[i])){    hash.Add(nums[i], i);    }    else{    if(target == nums[i] * 2){    return new int[2]{hash[nums[i]] + 1, i + 1};    }    // if one number appears twice and not equals to target half , just take the first one    }    }        foreach(var k in hash.Keys){    var k2 = target - k;    if(hash.ContainsKey(k2) && hash[k2] != hash[k]){    var index1 = hash[k];    var index2 = hash[k2];        if(index1 > index2){    return new int[2]{index2 + 1, index1 + 1};    }else{    return new int[2]{index1 + 1, index2 + 1};    }    }    }        return null;    }}