【leetcode系列】Two Sum

解法一，我自己想的，有点弱，时间复杂度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);
	}
}

解法二，网上大牛的，点击打开链接，很巧妙，用java的hashtable，典型的空间换时间，时间复杂度O(n)，但是要满足下标的要求，必须在找出满足条件的数字后再重新排个序，下面是我稍微修改的代码：

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;
}

【leetcode系列】Two Sum