LeetCode | 易学教程

Topic

Array
Hash Table

Description

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 1:

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

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

2 <= nums.length <= 10³
-10⁹ <= nums[i] <= 10⁹
-10⁹ <= target <= 10⁹
Only one valid answer exists.

Analysis

关键是利用HashMap做缓存，数组元素做键，下标作值。

Submission

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

public class TwoSum {
	public int[] twoSum(int[] nums, int target) {
		Map<Integer, Integer> map = new HashMap<>();

		for (int i = 0; i < nums.length; i++) {
			int diff = target - nums[i];
			if (map.containsKey(diff)) {
				return new int[] { map.get(diff), i };
			}
			map.put(nums[i], i);
		}
		return new int[] { -1, -1 };
	}
}

Test

import static org.junit.Assert.*;
import static org.hamcrest.collection.IsArrayContainingInAnyOrder.*;
import org.junit.Test;

public class TwoSumTest {

	@Test
	public void test() {
		TwoSum obj = new TwoSum();
		assertThat(toObject(obj.twoSum(new int[] { 2, 7, 11, 15 }, 9)), //
				arrayContainingInAnyOrder(0, 1));
		assertThat(toObject(obj.twoSum(new int[] { 3, 2, 4 }, 6)), //
				arrayContainingInAnyOrder(2, 1));
	}

	public static Integer[] toObject(int[] intArray) {
		Integer[] result = new Integer[intArray.length];
		for (int i = 0; i < intArray.length; i++) {
			result[i] = Integer.valueOf(intArray[i]);
		}
		return result;
	}
}

来源：oschina

链接：https://my.oschina.net/jallenkwong/blog/4836101

标签

java

junit