[LeetCode Java Solution] – sort/ two pointers/ merge- Merge k Sorted Lists

Problem:

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Method 1:

MergeSort. This is a very typical implementation for MergeSort by using array.

Code for method 1:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        return helper(lists, 0, lists.length-1);
    }
    
    public ListNode helper (ListNode[] lists, int l, int h){
        if (l < h){
             int mid = (l + h)/2;
             return merge(helper(lists, l, mid), helper(lists, mid+1, h));
        }
        return lists[l];
    }
    
    public ListNode merge(ListNode l1, ListNode l2){
        ListNode dummy = new ListNode(0);
        ListNode temp = dummy;
        while (l1 != null && l2 != null){
            if (l1.val < l2.val){
                temp.next = l1;
                temp = temp.next;
                l1 = l1.next;
            }else{
                temp.next = l2;
                temp = temp.next;
                l2 = l2.next;
            }
        }
        if (l1 != null){
            temp.next = l1;
        }
        if (l2 != null){
            temp.next = l2;
        }
        return dummy.next;
    }
}

Analysis for time complexity:

Suppose there are k lists, and there are n elements inside each list. The time complexity is O(nklogk).

Method 2:

Use heap. Suppose we have k lists, and for each list, there are at most n elements.

Maintain a list of k elements. And always remove the first element from the list. After removing an element, add the element after it to the heap.

Code for method 2:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<ListNode> heap = new PriorityQueue<ListNode>(10, new Comparator<ListNode>(){
            public int compare (ListNode l1, ListNode l2){
                return l1.val - l2.val;
            }
        });
        
        for (int i = 0; i < lists.length; i++){
            ListNode temp = lists[i];
            if (temp != null){
                heap.offer(temp);
            }
        }
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        
        while (heap.size() != 0){
            ListNode n = heap.poll();
            cur.next = n;
            cur = cur.next;
            if (n.next != null){
                heap.offer(n.next);
            }
        }
        return dummy.next;
    }
}

Analysis for time complexity:

We have to insert at most k*n nodes, and for each node to be in order in the heap, we need log(k) time (“Bottom Up Insertion”). So total time complexity is also knlog(k).

[LeetCode] – sort, merge – Merge Intervals

Problem:

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

Method 1:

Use the code for ‘Insert Intervals’.

Code for Method 1:

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public ArrayList<Interval> merge(List<Interval> intervals) {
        ArrayList<Interval> res = new ArrayList<Interval>();
        for (int i = 0; i < intervals.size(); i++){
            res = insert(res, intervals.get(i));
        }
        return res;
    }
    
    public ArrayList<Interval> insert(List<Interval> intervals, Interval newInterval){
        ArrayList<Interval> res = new ArrayList<Interval>();
        ArrayList<Interval> resb = new ArrayList<Interval>();
        if (intervals.size() == 0 ){
            res.add(newInterval);
            return res;
        }
        
        for (Interval i: intervals){
            if (i.end < newInterval.start) res.add(i);
            else if (newInterval.end < i.start) resb.add(i);
            else {
                newInterval.start = Math.min (i.start, newInterval.start);
                newInterval.end = Math.max (i.end, newInterval.end);
            }
        }
        res.add(newInterval);
        res.addAll(resb);
        return res;
    }
}

Method 2:

First, sort all the intervals according to their starting values. Then, add one into res. (We always keep last one in the res to be unsure. ) Every time, we get the last element of res, compare it with the current Interval in the set intervals. Basically only compare the “end” value.
Code for method 2:

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public ArrayList<Interval> merge(List<Interval> intervals) {
        ArrayList<Interval> res = new ArrayList<Interval>();
        if (intervals == null || intervals.size() == 0) return res;
        
        Comparator<Interval> comp = new Comparator<Interval>(){
            public int compare(Interval i1, Interval i2){
                if (i1.start > i2.start) return 1;
                else if (i1.start < i2.start) return -1;
                else return 0;
            }
        };
        Collections.sort(intervals, comp);
        
        res.add(intervals.get(0));
        
        for (int i = 1; i < intervals.size(); i++){
            if (res.get(res.size()-1).end >= intervals.get(i).start){
                res.get(res.size()-1).end = Math.max(res.get(res.size()-1).end, intervals.get(i).end);
            }else{
                res.add(intervals.get(i));
            }
        }
        return res;
    }
}

[LeetCode] -sort, merge – Insert Interval

Problem:

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

Method:

1. Iterate through the intervals set.

• If this.end < newInterval.start. No overlap, add this to res;
• If this.start > newInterval.end. No overlap, add this to res;
• If this.end >=newInterval.start || this.start <=newInterval.end. Merge and then add new interval to res.

Code:

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public ArrayList<Interval> insert(List<Interval> intervals, Interval newInterval) {
        ArrayList<Interval> res = new ArrayList<Interval>();
        ArrayList<Interval> resb = new ArrayList<Interval>();
        
        if (intervals.size() == 0) {
            res.add(newInterval);
            return res;
        }
        
        for (Interval i: intervals){
            if (i.end < newInterval.start) res.add(i);
            else if (i.start > newInterval.end) resb.add(i);
            else if (i.end >= newInterval.start || i.start <= newInterval.end){
                newInterval.start = Math.min(i.start,newInterval.start);
                newInterval.end = Math.max(i.end,newInterval.end);
            }
        }
        res.add(newInterval);
        res.addAll(resb);
        
        return res;
    }
}

[LeetCode] – Two pointers / Merge- Merge Sorted Array

Problem:

Given two sorted integer arrays A and B, merge B into A as one sorted array.

Note:
You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m and n respectively.

Method:

Similar as the problem : ‘Merge sorted list’. However, the difference is: we start from the end, and work backwards to compare elements.

Code:

public class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1;
        int j = n - 1;
        int k = m + n -1;

        while (i >= 0 && j >= 0){
            if (nums1[i] > nums2[j]){
                nums1[k] = nums1[i];
                i--;
                k--;
            }else{
                nums1[k] = nums2[j];
                j--;
                k--;
            }
        }
        
        while(i >= 0){
            nums1[k] = nums1[i];
            i--;
            k--;
        }
        while(j >= 0){
            nums1[k] = nums2[j];
            j--;
            k--;
        }
    }
}

To make use of the property that this will be an in-place operation for nums1, we can simplify the code to be:

public class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m-1;
        int j = n-1;
        int k = m+n-1;
        
        while (k >= 0){
            if (j < 0 || (i >= 0 && nums1[i] > nums2[j])){
                nums1[k--] = nums1[i--];
            }else{
                nums1[k--] = nums2[j--];
            }
        }
    }
}

[LeetCode] – Sort/ Two pointers/Merge – Merge Two Sorted List

Problem:

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Method 1:

1. Use two pointers to move through the two lists. Compare and link the smaller one to the result list.

2. Use a dummy node to avoid choosing head for the new list.

3. To save space, do not have to create new node.

Code for method 1:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) return null;
   
        ListNode dummy = new ListNode(1);
        ListNode temp0 = dummy;
        ListNode temp1 = l1;
        ListNode temp2 = l2;
        
        while (temp1 != null && temp2 != null){
            if (temp1.val <= temp2.val){
              //  ListNode item = new ListNode(temp1.val);
              // to save space, do not have to create new object
                temp0.next = temp1;
                temp1 = temp1.next;
                temp0 = temp0.next;
            }else{
            //    ListNode item = new ListNode(temp2.val);
                temp0.next = temp2;
                temp2 = temp2.next;
                temp0 = temp0.next;
            }
        }
        if (temp1 != null){
            temp0.next = temp1;
        }
        if (temp2 != null){
            temp0.next = temp2;
        }
        return dummy.next;
    }
}