Two Sum

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

代码：

import java.util.HashMap;

public class TwoSum {

    public static void main(String[] args) {
        int[] sums = { 3, 2, 4 };
        int target = 6;
        int[] res = twoSum(sums, target);
        int[] indexs = getTwoIndexs(sums,target);
        for (int i = 0; i < res.length; i++) {
            System.out.println(res[i]);
        }
        for (int i = 0; i < indexs.length; i++) {
            System.out.println(indexs[i]);
        }
    }

    // HashMap的解决方式，参考了网上的，又学习了HashMap
    private static int[] getTwoIndexs(int[] nums, int target) {
        int[] twoIndexs = new int[2];
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i])) {
                int index = map.get(nums[i]);
                twoIndexs[0] = Math.min(i+1, index+1);
                twoIndexs[1] = Math.max(i+1, index+1);
                break;
            }
            int value = target - nums[i];
            map.put(value, i);
        }
        return twoIndexs;
    }

    // 之前自己写的解决方式，明显不好
    private static int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    res[0] = i + 1;
                    res[1] = j + 1;
                }
            }
        }
        return res;
    }

}