18. 4Sum

The basic idea is turning 4Sum into 3Sum. In order to improve efficiency, there are some tricks to jump out of current traversal:

1. duplicate removal
2. current num is too small
3. current num is too large

The details are shown is the following code with comment:

public List<List<Integer>> fourSum(int[] nums, int target) {
    Arrays.sort(nums);
    List<List<Integer>> resultList = new ArrayList<List<Integer>>();
    for(int i = 0; i < nums.length - 3; i++) {
        if( i > 0 && nums[i] == nums[i - 1] ) // remove duplicate
            continue;
        if( nums[i] + 3 * nums[nums.length - 1] < target ) // current num is too small
            continue;
        if( nums[i] * 4 > target ) // current num is too large
            break;
        for(int j = i + 1; j < nums.length - 2; j++) {
            if( j > i + 1 && nums[j] == nums[j - 1] ) // duplicate removal
                continue;
            if( nums[i] + nums[j] + 2 * nums[nums.length - 1] < target ) // current num is too small
                continue;
            if( nums[i] + nums[j] * 3 > target ) // current num is too large
                break;
            int begin = j + 1, end = nums.length - 1;
            while( begin < end ) {
                int sum = nums[i] + nums[j] + nums[begin] + nums[end];
                if( sum > target ) 
                    end--;
                else if( sum < target )
                    begin++;
                else if( sum == target ) {
                    if( begin == j + 1 || nums[begin] != nums[begin - 1] ) {
                        List<Integer> currentList = new ArrayList<Integer>();
                        currentList.add(nums[i]);
                        currentList.add(nums[j]);
                        currentList.add(nums[begin]);
                        currentList.add(nums[end]);
                        resultList.add(currentList);
                    }
                    begin++;
                    end--;
                }
            }
        }
    }
    return resultList;
}