[LeetCode] 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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution 1:
Approach: Brute force
Time complexity: O(n^2)
Space complexity: O(1)

public class Solution {
    public int[] twoSum(int[] nums, int target) {
    
        int[] result=new int[2];
        if(nums==null||nums.length==0) return result;
        for(int i=0; i<nums.length; i++){
            for(int j=i+1; j<nums.length; j++){
                if(nums[i]+nums[j]==target){
                    result[0]=i;
                    result[1]=j;
                }
            }
        }
        return result;       
    }
}

Solution 2:
Approach: Hash Table
Time Complexity: O(n)
Space Complexity: O(n)

public class Solution {
    public int[] twoSum(int[] nums, int target) {
       
        int[] result=new int[2];
        if(nums==null||nums.length==0) return result;
        HashMap<Integer, Integer> hashMap=new HashMap<Integer, Integer>();
        for(int i=0; i<nums.length; i++){
            if(hashMap.containsKey(target-nums[i])){
                result[0]=i;
                result[1]=hashMap.get(target-nums[i]);
            }else{
                hashMap.put(nums[i], i);
            }
        }
        return result;
    }
}