TwoSum

  给一个数组，还有一个target，找到数组中能够产生和值为target的两个数，返回2个数的index。 假设每次输入，肯定有唯一解
  2次for循环得出一个暴力解，但是超时了。

  或者空间换时间，把每一个数对应需要的另一个值保存在hashmap中，然后再来一次循环，如果包含该key，就对了

“`java
/*
* Given an array of integers, find two numbers such that they add up to a
* specific target number.
*
* The function twoSum should return indices of the two numbers such that
* they add up to the target, where index1 must be less than index2. Please
* note that your returned answers (both index1 and index2) are not
* zero-based.
*
* You may assume that each input would have exactly one solution.
*
* Input: numbers={2, 7, 11, 15}, target=9 Output: index1=1, index2=2
* 给一个数组，还有一个target，找到数组中能够产生和值为target的两个数，返回2个数的index。 假设每次输入，肯定有唯一解
* 2次for循环得出一个暴力解，但是超时了。
*
* 或者空间换时间，把每一个数对应需要的另一个值保存在hashmap中，然后再来一次循环，如果包含该key，就对了
*/

import java.util.HashMap;

public class TwoSum {

public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    HashMap<Integer, Integer> another = new HashMap<Integer, Integer>();
    for (int i = 0; i < numbers.length; i++) {
        another.put(target - numbers[i], i);
    }
    for (int i = 0; i < numbers.length; i++) {
        if (another.containsKey(numbers[i])) {
            if (another.get(numbers[i]) != i) {
                result[0] = i > another.get(numbers[i]) ? another
                        .get(numbers[i]) + 1 : i + 1;
                result[1] = i < another.get(numbers[i]) ? another
                        .get(numbers[i]) + 1 : i + 1;
                return result;
            }
        }
    }
    return null;
}

public static void main(String[] args) {
    // TODO Auto-generated method stub

}

}