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.

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)

找出一個(gè)序列中所有和為target的四元組。
循環(huán)的3Sum。

public class Solution {
  public List<List<Integer>> fourSum(int[] nums, int target) {
    List result = new ArrayList<List<Integer>>();
    if (nums == null || nums.length < 4) return result;
    
    Arrays.sort(nums);
    for (int i = 0; i < nums.length - 3; i++) {
        
        if (i > 0 && nums[i] == nums[i-1]) continue;
        
        for (int j = i + 1; j < nums.length - 2; j++) {
            
            if (j > i + 1 && nums[j] == nums[j-1]) continue;
            
            int left = j + 1;
            int right = nums.length - 1;
            while (left < right) {
                int tmp = nums[i] + nums[j] + nums[left] + nums[right];
                if (tmp == target) {
                    List l = new ArrayList<Integer>();
                    l.add(nums[i]);
                    l.add(nums[j]);
                    l.add(nums[left]);
                    l.add(nums[right]);
                    result.add(l);
                    while (left < right && nums[++left] == nums[left-1]);
                    while (left < right && nums[--right] == nums[right+1]);
                } else if (tmp > target) while (left < right && nums[--right] == nums[right+1]);
                else while (left < right && nums[++left] == nums[left-1]);
            }
        }
    }
    return result;
  }
}