LeetCode --- 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, and you may not use the same element twice.

Example:

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

Approach 1

public class TwoSums {

    @Test
    public void test () {
        Assert.assertArrayEquals(new int[] {0,1}, twoSum(new int[]{2,7,11,15}, 9));
        Assert.assertArrayEquals(new int[] {1,2}, twoSum(new int[]{3,2,4}, 6));
    }

    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] == target - nums[j]) {
                    return new int[] {i, j};
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }

}

Approach 2

public class TwoSums {

    @Test
    public void test () {
        Assert.assertArrayEquals(new int[] {0,1}, twoSum(new int[]{2,7,11,15}, 9));
        Assert.assertArrayEquals(new int[] {1,2}, twoSum(new int[]{3,2,4}, 6));
    }

    public int[] twoSum(int[] nums, int target) {
        AtomicInteger integer = new AtomicInteger(0);
        Map<Integer, Integer> collect = Arrays.stream(nums).boxed().collect(Collectors.toMap(Function.identity(), o -> integer.getAndIncrement()));

        for (Map.Entry<Integer, Integer> entry : collect.entrySet()) {
            Integer first = entry.getKey();
            if (collect.containsKey(target - first)) {
                return new int[] {collect.get(first),collect.get(target-first)};
            }
        }

        throw new IllegalArgumentException("No two sum solution");
    }

}