1. Two Sum_Easy(eg: [3,2,4], target, 6)

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example One:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Solution: 

class Solution {
     public int[] twoSum(int[] nums, int target) {

     HashMap<Integer,Integer> map = new HashMap<>();
            int j = 0;
        for(Integer k : nums){
            map.put(k,j);
            j++;
        }

        for(int i = 0; i < nums.length; i++){
            int val = target - nums[i];
            if(map.containsKey(val) && map.get(val) != i) return new int[] {i, map.get(val)};
        }
        return null;
        
    }
}

Hints:

Logics:

• Has a HashMap to store the vales in the array
• put the values as the key and the order as the vlaue in the map
• iterate the array nums, and find the value after minus
• don't forget the special case : if the 3 is 0, and also 3 is also in the map, and the target is 6, but the answer is wrong,because there is only one 3 not two