leetcode18. 4Sum

題目

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]
]

思路

在原來3sum的基礎上面加一層循環

代碼

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();

        if(nums.length < 4){
            return res;
        }
        Arrays.sort(nums);

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

        int t;
        for( int i = 0; i < len - 3; i ++){
            t = nums[i];
            if(i > 0 && t == nums[i - 1])
                continue;
            if(4 * t > target)
                break;
            if(4 * t == target && i < len - 3&& nums[i + 3] == t){
                res.add(Arrays.asList(t, t, t, t));
                continue;
            }   
            threeSum(nums, target - t, i+1, len, res, t);
        }
        return res;
    }

    public void  threeSum(int[] nums, int target, int lo, int hi, List<List<Integer>> res, int t) {

 for(int i  = lo; i <  hi- 2; i++){
            if(i == lo || ( nums[i] != nums[i-1])){
                int low = i+1;int high = nums.length - 1; int iSum = target  - nums[i];
                while(low < high){
                    if(nums[low] + nums[high] == iSum){
                        res.add(Arrays.asList(t, nums[i], nums[low], nums[high]));
                        while(low < high && nums[low] == nums[low + 1]) low++;
                        while(low < high && nums[high] == nums[high - 1]) high--;
                        low++; high--;
                    }
                    else if( nums[low] + nums[high] > iSum){
                        high--;
                    }
                    else{
                        low++;
                    }
                }
            }
        }
    }
}

他山之玉

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;
    }