Merging K Sorted Linked Lists











up vote
0
down vote

favorite












Specifically, how can I improve the time complexity of my algorithm (currently it is O(listLength * numberOfLists))? It only beats 5% of accepted LeetCode solutions, which surprised me.



/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
private void advance(final ListNode listNodes, final int index) {
listNodes[index] = listNodes[index].next;
}

public ListNode mergeKLists(final ListNode listNodes) {
ListNode sortedListHead = null;
ListNode sortedListNode = null;

int associatedIndex;

do {
int minValue = Integer.MAX_VALUE;
associatedIndex = -1;

for (int listIndex = 0; listIndex < listNodes.length; listIndex++) {
final ListNode listNode = listNodes[listIndex];

if (listNode != null && listNode.val < minValue) {
minValue = listNode.val;
associatedIndex = listIndex;
}
}

// An associated index of -1 indicates no more values left in any of the given lists
if (associatedIndex != -1) {
if (sortedListNode == null) {
sortedListNode = new ListNode(minValue);
sortedListHead = sortedListNode;
}
else {
sortedListNode.next = new ListNode(minValue);
sortedListNode = sortedListNode.next;
}

advance(listNodes, associatedIndex);
}
}
while (associatedIndex != -1);

return sortedListHead;
}
}


Note that the Solution class in addition to ListNode is already provided, the only code that I wrote was inside mergeKLists.










share|improve this question


























    up vote
    0
    down vote

    favorite












    Specifically, how can I improve the time complexity of my algorithm (currently it is O(listLength * numberOfLists))? It only beats 5% of accepted LeetCode solutions, which surprised me.



    /**
    * Definition for singly-linked list.
    * public class ListNode {
    * int val;
    * ListNode next;
    * ListNode(int x) { val = x; }
    * }
    */
    class Solution {
    private void advance(final ListNode listNodes, final int index) {
    listNodes[index] = listNodes[index].next;
    }

    public ListNode mergeKLists(final ListNode listNodes) {
    ListNode sortedListHead = null;
    ListNode sortedListNode = null;

    int associatedIndex;

    do {
    int minValue = Integer.MAX_VALUE;
    associatedIndex = -1;

    for (int listIndex = 0; listIndex < listNodes.length; listIndex++) {
    final ListNode listNode = listNodes[listIndex];

    if (listNode != null && listNode.val < minValue) {
    minValue = listNode.val;
    associatedIndex = listIndex;
    }
    }

    // An associated index of -1 indicates no more values left in any of the given lists
    if (associatedIndex != -1) {
    if (sortedListNode == null) {
    sortedListNode = new ListNode(minValue);
    sortedListHead = sortedListNode;
    }
    else {
    sortedListNode.next = new ListNode(minValue);
    sortedListNode = sortedListNode.next;
    }

    advance(listNodes, associatedIndex);
    }
    }
    while (associatedIndex != -1);

    return sortedListHead;
    }
    }


    Note that the Solution class in addition to ListNode is already provided, the only code that I wrote was inside mergeKLists.










    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Specifically, how can I improve the time complexity of my algorithm (currently it is O(listLength * numberOfLists))? It only beats 5% of accepted LeetCode solutions, which surprised me.



      /**
      * Definition for singly-linked list.
      * public class ListNode {
      * int val;
      * ListNode next;
      * ListNode(int x) { val = x; }
      * }
      */
      class Solution {
      private void advance(final ListNode listNodes, final int index) {
      listNodes[index] = listNodes[index].next;
      }

      public ListNode mergeKLists(final ListNode listNodes) {
      ListNode sortedListHead = null;
      ListNode sortedListNode = null;

      int associatedIndex;

      do {
      int minValue = Integer.MAX_VALUE;
      associatedIndex = -1;

      for (int listIndex = 0; listIndex < listNodes.length; listIndex++) {
      final ListNode listNode = listNodes[listIndex];

      if (listNode != null && listNode.val < minValue) {
      minValue = listNode.val;
      associatedIndex = listIndex;
      }
      }

      // An associated index of -1 indicates no more values left in any of the given lists
      if (associatedIndex != -1) {
      if (sortedListNode == null) {
      sortedListNode = new ListNode(minValue);
      sortedListHead = sortedListNode;
      }
      else {
      sortedListNode.next = new ListNode(minValue);
      sortedListNode = sortedListNode.next;
      }

      advance(listNodes, associatedIndex);
      }
      }
      while (associatedIndex != -1);

      return sortedListHead;
      }
      }


      Note that the Solution class in addition to ListNode is already provided, the only code that I wrote was inside mergeKLists.










      share|improve this question













      Specifically, how can I improve the time complexity of my algorithm (currently it is O(listLength * numberOfLists))? It only beats 5% of accepted LeetCode solutions, which surprised me.



      /**
      * Definition for singly-linked list.
      * public class ListNode {
      * int val;
      * ListNode next;
      * ListNode(int x) { val = x; }
      * }
      */
      class Solution {
      private void advance(final ListNode listNodes, final int index) {
      listNodes[index] = listNodes[index].next;
      }

      public ListNode mergeKLists(final ListNode listNodes) {
      ListNode sortedListHead = null;
      ListNode sortedListNode = null;

      int associatedIndex;

      do {
      int minValue = Integer.MAX_VALUE;
      associatedIndex = -1;

      for (int listIndex = 0; listIndex < listNodes.length; listIndex++) {
      final ListNode listNode = listNodes[listIndex];

      if (listNode != null && listNode.val < minValue) {
      minValue = listNode.val;
      associatedIndex = listIndex;
      }
      }

      // An associated index of -1 indicates no more values left in any of the given lists
      if (associatedIndex != -1) {
      if (sortedListNode == null) {
      sortedListNode = new ListNode(minValue);
      sortedListHead = sortedListNode;
      }
      else {
      sortedListNode.next = new ListNode(minValue);
      sortedListNode = sortedListNode.next;
      }

      advance(listNodes, associatedIndex);
      }
      }
      while (associatedIndex != -1);

      return sortedListHead;
      }
      }


      Note that the Solution class in addition to ListNode is already provided, the only code that I wrote was inside mergeKLists.







      java algorithm linked-list






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 11 mins ago









      Mar Dev

      26229




      26229



























          active

          oldest

          votes











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ifUsing("editor", function () {
          StackExchange.using("externalEditor", function () {
          StackExchange.using("snippets", function () {
          StackExchange.snippets.init();
          });
          });
          }, "code-snippets");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "196"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f208315%2fmerging-k-sorted-linked-lists%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown






























          active

          oldest

          votes













          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f208315%2fmerging-k-sorted-linked-lists%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Feedback on college project

          Futebolista

          Albești (Vaslui)