leetcode:Two Sum (java)

package Leetcode;
/**
 * 题目：
 *      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].
 *
 * 解题思路：
 *      先遍历一个数字，吧另外一个数字用一个HashMap存储起来，建立数字和其坐标位置之间的映射，（HashMap是常数级的查找效率），
 *      在遍历数组的时候，用target减去遍历到的数字，就是另一个需要的数字了，直接在HashMap中查找其是否存在即可。
 *      注意：要判断查找到的数字不是第一个数字，比如target是4，遍历到了一个2，那么另外一个2不能是之前那个2，
 */

import java.util.HashMap;

public class TwoSum_1008 {
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        //判断是否为合法输入
        if (nums == null || nums.length < 2) {
            return result;
        }
        HashMap<Integer, Integer> map = new HashMap<>();

        //遍历数组
        for (int i = 0; i < nums.length; i++) {
            int temp = target - nums[i];

            //判断HashMap中是否存在要找的数字
            if (map.containsKey(temp) && map.get(temp) != i) {
                result[0] = map.get(temp);
                result[1] = i;

                break;
            }
            else
                map.put(nums[i], i);
        }
        return result;
    }

    public static void main(String[] args) {
        int[] nums = {2, 7, 11, 15};
        int target = 9;

        TwoSum_1008 test = new TwoSum_1008();
        int[] result = new int[2];
        result = test.twoSum(nums, target);
        System.out.println(result[0] + " " + result[1]);
    }
}