LeetCode_1. Two Sum

 

1. Two Sum

Easy

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].

package leetcode.easy;

import java.util.HashMap;
import java.util.Map;

public class TwoSum {
	@org.junit.Test
	public void test() {
		int[] nums = { 2, 7, 11, 15 };
		int target = 9;
		TwoSum twoSum = new TwoSum();
		int[] result = twoSum.twoSum1(nums, target);
		System.out.println(result[0] + " " + result[1]);

		result = twoSum.twoSum2(nums, target);
		System.out.println(result[0] + " " + result[1]);

		result = twoSum.twoSum3(nums, target);
		System.out.println(result[0] + " " + result[1]);
	}

	public int[] twoSum1(int[] nums, int target) {
		for (int i = 0; i < nums.length; i++) {
			for (int j = i + 1; j < nums.length; j++) {
				if (nums[j] == target - nums[i]) {
					return new int[] { i, j };
				}
			}
		}
		throw new IllegalArgumentException("No two sum solution");
	}

	public int[] twoSum2(int[] nums, int target) {
		Map<Integer, Integer> map = new HashMap<>();
		for (int i = 0; i < nums.length; i++) {
			map.put(nums[i], i);
		}
		for (int i = 0; i < nums.length; i++) {
			int complement = target - nums[i];
			if (map.containsKey(complement) && map.get(complement) != i) {
				return new int[] { i, map.get(complement) };
			}
		}
		throw new IllegalArgumentException("No two sum solution");
	}

	public int[] twoSum3(int[] nums, int target) {
		Map<Integer, Integer> map = new HashMap<>();
		for (int i = 0; i < nums.length; i++) {
			int complement = target - nums[i];
			if (map.containsKey(complement)) {
				return new int[] { map.get(complement), i };
			}
			map.put(nums[i], i);
		}
		throw new IllegalArgumentException("No two sum solution");
	}
}