[leetcode]Two Sum

本来以为是一道水题，结果居然能从n^2, n*log(n)一直优化到n。感慨啊~

O(n^2): 暴力

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
    	int[] r = 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)
        		{
        			r[0] = i+1;
        			r[1] = j+1;
        			break;
        		}
        	}
        }
        return r;
    }
}

O(n*log(n)): 排序。用了O(n)的空间记录原来的数组，然后开始和最后O(n)的扫一遍，不过不改变总体复杂度。当然有的人用数据结构记录原来的和后来的联系，那也就是另外O(n)的空间省最后扫一遍而已。解题过程中一开始忘了记录原来的数组，后来忘了排序后的第一第二和排序后的也可能不同。

import java.util.Arrays;

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int[] result = new int[2];
    	int[] tmp = new int[numbers.length];
    	for (int i = 0; i < numbers.length; i++)
    	{
    		tmp[i] = numbers[i];
    	}
        Arrays.sort(tmp);
        int l = 0;
        int r = tmp.length - 1;
        int sum = tmp[l] + tmp[r];
        while( sum != target && r > l)
        {
        	if (sum > target)
        	{
        		r--;
        	}
        	else
        	{
        		l++;
        	}
        	sum = tmp[l] + tmp[r];
        }
        int index = 0;
        for (int i = 0; i < numbers.length; i++)
        {
        	if (tmp[l] == numbers[i] || tmp[r] == numbers[i])
        	{
        		result[index] = i + 1;
        		index++;
        		if (index > 1) break;
        	}
        }
        
        return result;
    }
}

　O(n): 果然无耻的用了Hash，那么也就不神奇了。所以真实的复杂度考虑到Hash的实现，会比O(n)要大些。

import java.util.HashMap;

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int[] result = new int[2];
    	HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    	for (int i = 0; i < numbers.length; i++)
    	{
    		if (map.containsKey(target - numbers[i]))
    		{
    			result[0] = map.get(target-numbers[i]) + 1;
    			result[1] = i + 1;
    			break;
    		}
    		else
    		{
    			map.put(numbers[i], i);
    		}
    	}
        
        return result;
    }
}