18 4Sum

题目

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: The solution set must not contain duplicate quadruplets.

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]
]

Code

public List<List<Integer>> fourSum(int[] nums, int target) {
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        int len = nums.length;
        if (nums == null || len < 4)
            return res;

        Arrays.sort(nums);

        int max = nums[len - 1];
        if (4 * nums[0] > target || 4 * max < target)
            return res;

        int i, z;
        for (i = 0; i < len; i++) {
            z = nums[i];
            if (i > 0 && z == nums[i - 1])// avoid duplicate
                continue;
            if (z + 3 * max < target) // z is too small
                continue;
            if (4 * z > target) // z is too large
                break;
            if (4 * z == target) { // z is the boundary
                if (i + 3 < len && nums[i + 3] == z)
                    res.add(Arrays.asList(z, z, z, z));
                break;
            }

            threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);
        }

        return res;
    }

    /*
     * Find all possible distinguished three numbers adding up to the target
     * in sorted array nums[] between indices low and high. If there are,
     * add all of them into the ArrayList fourSumList, using
     * fourSumList.add(Arrays.asList(z1, the three numbers))
     */
    public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
            int z1) {
        if (low + 1 >= high)
            return;

        int max = nums[high];
        if (3 * nums[low] > target || 3 * max < target)
            return;

        int i, z;
        for (i = low; i < high - 1; i++) {
            z = nums[i];
            if (i > low && z == nums[i - 1]) // avoid duplicate
                continue;
            if (z + 2 * max < target) // z is too small
                continue;

            if (3 * z > target) // z is too large
                break;

            if (3 * z == target) { // z is the boundary
                if (i + 1 < high && nums[i + 2] == z)
                    fourSumList.add(Arrays.asList(z1, z, z, z));
                break;
            }

            twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);
        }

    }

    /*
     * Find all possible distinguished two numbers adding up to the target
     * in sorted array nums[] between indices low and high. If there are,
     * add all of them into the ArrayList fourSumList, using
     * fourSumList.add(Arrays.asList(z1, z2, the two numbers))
     */
    public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
            int z1, int z2) {

        if (low >= high)
            return;

        if (2 * nums[low] > target || 2 * nums[high] < target)
            return;

        int i = low, j = high, sum, x;
        while (i < j) {
            sum = nums[i] + nums[j];
            if (sum == target) {
                fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));

                x = nums[i];
                while (++i < j && x == nums[i]) // avoid duplicate
                    ;
                x = nums[j];
                while (i < --j && x == nums[j]) // avoid duplicate
                    ;
            }
            if (sum < target)
                i++;
            if (sum > target)
                j--;
        }
        return;
    }

MyCode

class Solution {
    public List<List<Integer>> fourSum(int[] num, int target) {
        Arrays.sort( num );
        List<List<Integer>> rt = new LinkedList<>();
        int sum = 0;

        for ( int i = 0 ;i< num.length -3;i++){
            if( i == 0 || (i> 0 && num[i] != num[i-1]) )
            for( int j = i + 1;j< num.length - 2 ; j++){
                if( j == 1 || j==i+1 && num[j]== num[i]||(j>1 &&num[j]!=num[j-1])){
                    sum = target - num[i] - num[j];
                    int lo = j + 1;
                    int hi = num.length-1;
                    while( lo < hi ){
                        if( num[lo] + num[hi] == sum ){
                            rt.add(( Arrays.asList(num[i],num[j],num[lo],num[hi])));
                            while( lo < hi && num[hi]==num[hi-1]) hi--;
                            while( lo < hi && num[lo]==num[lo+1]) lo++;
                            lo++;hi--;
                        }else if( num[lo] + num[hi] < sum)
                            lo++;
                        else hi--;
                    }
                }
            }
        }
        return rt;
    }
}

问题

完全是仿照3sum的思路写的，但是思路更加复杂，最后的逻辑可以说是凑出来的，并不能完全理通顺（Mycode）