[Leetcode series] Two sum

Solution 1: what I think is a little weak, time complexity O (N ^ 2 ).

public class Solution {public int[] twoSum(int[] numbers, int target) {int[] result = new int[2];for (int i = 0; i < numbers.length; i++) {for (int j = i + 1; j < numbers.length; j++) {if (numbers[i] + numbers[j] == target) {result[0] = i;result[1] = j;}}}if (result[0] == 0) {if (result[1] != 1) {int tmp = numbers[0];numbers[0] = numbers[1];numbers[1] = tmp;result[0] = 1;} else {int tmp = numbers[1];numbers[1] = numbers[2];numbers[2] = tmp;result[1] = 2;tmp = numbers[0];numbers[0] = numbers[1];numbers[1] = tmp;result[0] = 1;}}return result;}public void arrayPrint(int[] array) {for (int i = 0; i < array.length; i++) {System.out.print(array[i] + " ");}System.out.println();}public static void main(String[] args) {int[] numbers = { 11, -22, 12, 7, 2 };int target = 9;Solution mSolution = new Solution();mSolution.arrayPrint(numbers);int[] result = mSolution.twoSum(numbers, target);mSolution.arrayPrint(result);mSolution.arrayPrint(numbers);}}

Solution 2: Click to open the link on the Internet. It is clever to use Java hashtable to change the typical space for time and time complexity O (N), but it must meet the requirements of the following standards, after finding the numbers that meet the conditions, you must sort them again. The following code is slightly modified:

public int[] twoSum_2(int[] numbers, int target) {int[] result = new int[2];HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();for(int i = 0; i < numbers.length; i++) {if(hm.get(target - numbers[i]) != null) {result[0] = hm.get(target - numbers[i]);result[1] = i;} else {hm.put(numbers[i], i);}}if (result[0] == 0) {if (result[1] != 1) {int tmp = numbers[0];numbers[0] = numbers[1];numbers[1] = tmp;result[0] = 1;} else {int tmp = numbers[1];numbers[1] = numbers[2];numbers[2] = tmp;result[1] = 2;tmp = numbers[0];numbers[0] = numbers[1];numbers[1] = tmp;result[0] = 1;}}return result;}