K Sum Problem

All K Sum problem can be divided into two problems:

  • 2 Sum Problem
  • Reduce K Sum problem to K – 1 Sum Problem

从4Sum和3Sum,我们可以看出对于KSum的通用套路:将KSum转化为K-1 Sum,最后用2Sum的Two Pointer求解。

注意要点:先排序去除/跳过重复元素

Generalized K Sum Approach:

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        return kSum(nums, 0, 4, target);
    }

    private List<List<Integer>> kSum (int[] nums, int start, int k, int target) {
        int len = nums.length;
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (k == 2) { 
            //two pointers from left and right
            int left = start, right = len - 1;
            while (left < right) {
                int sum = nums[left] + nums[right];
                if (sum == target) {
                    List<Integer> path = new ArrayList<Integer>();
                    path.add(nums[left]);
                    path.add(nums[right]);
                    res.add(path);
                    left++;
                    right--;

                    while(left < right && nums[left] == nums[left - 1]) left++;
                    while(left < right && nums[right] == nums[right + 1]) right--;

                } else if (sum < target) { //move left
                    left++;
                } else { //move right
                    right--;
                }
            }
        } else {
            for(int i = start; i < len - (k - 1); i++) {
                if(i > start && nums[i] == nums[i - 1]) continue;
                List<List<Integer>> temp = kSum(nums, i + 1, k - 1, target - nums[i]);
                for(List<Integer> t : temp) {
                    t.add(0, nums[i]);
                }                    
                res.addAll(temp);
            }
        }
        return res;
    }
}

Another K Sum

public class Solution {
    int len = 0;
    public List<List<Integer>> fourSum(int[] nums, int target) {
        len = nums.length;
        Arrays.sort(nums);
        return kSum(nums, target, 4, 0);
    }
    private ArrayList<List<Integer>> kSum(int[] nums, int target, int k, int index) {
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        if (index >= len) {
            return res;
        }
        if (k == 2) {
            int i = index, j = len - 1;
            while (i < j) {
                //find a pair
                if (target - nums[i] == nums[j]) {
                    List<Integer> temp = new ArrayList<>();
                    temp.add(nums[i]);
                    temp.add(target-nums[i]);
                    res.add(temp);
                    i++;
                    j--;
                    //skip duplication
                    while (i < j && nums[i] == nums[i - 1]) i++;
                    while (i < j && nums[j] == nums[j + 1]) j--;
                } else if (target - nums[i] > nums[j]) {
                    //move left bound
                    i++;
                } else {
                    //move right bound
                    j--;
                }
            }
        } else {
            for (int i = index; i < len - k + 1; i++) {
                // Alternative: Skipping duplicates in the front
                // if (i > index && nums[i] == nums[i - 1]) {
                //    continue;
                // }
                //use current number to reduce k sum into k-1 sum
                ArrayList<List<Integer>> temp = kSum(nums, target - nums[i], k - 1, i + 1);
                if (temp != null) {
                    //add previous results
                    for (List<Integer> t : temp) {
                        t.add(0, nums[i]);
                    }
                    res.addAll(temp);
                }
                //skip duplicated numbers
                while (i < len - 1 && nums[i] == nums[i + 1]) {
                    i++;
                }
            }
        }
        return res;
    }
}

Reference

Generalized KSums: https://leetcode.com/problems/4sum/discuss/8609/My-solution-generalized-for-kSums-in-JAVA

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