58.4Sum-四数之和（中等题）

四数之和

1. 题目

   给一个包含n个数的整数数组S，在S中找到所有使得和为给定整数target的四元组(a, b, c, d)。

   注意事项
   四元组(a, b, c, d)中，需要满足a <= b <= c <= d
   答案中不可以包含重复的四元组。

2. 样例

   例如，对于给定的整数数组S=[1, 0, -1, 0, -2, 2] 和 target=0. 满足要求的四元组集合为：
   (-1, 0, 0, 1)
   (-2, -1, 1, 2)
   (-2, 0, 0, 2)

3. 题解

57.3Sum-三数之和（中等题）的难度升级版，思路完全一致。

public class Solution {
    /**
     * @param numbers : Give an array numbersbers of n integer
     * @param target : you need to find four elements that's sum of target
     * @return : Find all unique quadruplets in the array which gives the sum of
     *           zero.
     */
    public ArrayList<ArrayList<Integer>> fourSum(int[] numbers, int target) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        Arrays.sort(numbers);

        for (int i=0;i<numbers.length-3;i++)
        {
            if (i!=0 && numbers[i] ==numbers[i-1])
            {
                continue;
            }
            for (int j=i+1;j<numbers.length-2;j++)
            {
                if (j != i+1 && numbers[j] ==numbers[j-1])
                {
                    continue;
                }
                int start = j+1;
                int end = numbers.length-1;
                while (start < end)
                {
                    int sum = numbers[i] + numbers[j] + numbers[start] + numbers[end];
                    if (sum == target)
                    {
                        ArrayList<Integer> list = new ArrayList<Integer>();
                        list.add(numbers[i]);
                        list.add(numbers[j]);
                        list.add(numbers[start]);
                        list.add(numbers[end]);
                        result.add(list);
                        start++;
                        end--;
                        while (start < end && numbers[start] == numbers[start-1])
                        {
                            start++;
                        }
                        while (start < end && numbers[end] == numbers[end+1])
                        {
                            end--;
                        }
                    }
                    else if (sum < target)
                    {
                        start++;
                    }
                    else
                    {
                        end--;
                    }
                }

            }
        }

        return result;
    }
}

Last Update 2016.9.30