4 Sum

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Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)

The solution set must not contain duplicate quadruplets.

Example

For example, given array S = {1 0 -1 0 -2 2}, and target = 0. A solution set is:

(-1, 0, 0, 1)

(-2, -1, 1, 2)

(-2, 0, 0, 2)

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>> rst = new ArrayList<ArrayList<Integer>>();
        if (numbers == null || numbers.length == 0) {
            return rst;
        }
        Arrays.sort(numbers);
        helper(numbers, 0, target, 0, 0, new ArrayList<Integer>(), rst);
        return rst;
    }
    void helper(int[] numbers, int pos, int target, int sum, int level, 
    ArrayList<Integer> cur, ArrayList<ArrayList<Integer>> rst){
        if (level == 4) {
            if (sum == target) {
                rst.add(new ArrayList(cur));
            }
            return;
        }
        for (int i = pos; <span style="color:#ff6666;">i < numbers.length</span>/** && sum + numbers[i] <= target*/; i++) {
            if (i > 0 && i > pos && numbers[i] == numbers[i - 1]) {
                continue;
            }
            cur.add (numbers[i]);
            helper (numbers, i + 1, target, sum + numbers[i], level + 1, cur, rst);
            cur.remove(cur.size() - 1);
        }
    }
}

最開始紅色部分的循環條件我寫的是  for (int i = pos; i < numbers.length && sum + numbers[i] <= target; i++)

這樣寫不對，因爲numbers可能全是負數，比如 [-101, -1, -2, -3] 如果target是-107的話我的寫法就一個4 sum都找不出來了。