(Algorithms in Javascript) Leetcode 92. Reverse Linked List II - Reverse a linked list from position m to n. Do it in one-pass.

92-leetcode-reverse-linked-list-ii.js

const reverseBetween = (head, m, n) => {
  if(m>=n) return head;
  let currentNode = head;
  let previousNode = head;
  let temp = null, temp2 = null, tempStart = null;
  let countm = 0, countn=0;
  if(!currentNode.next) return head;
  
  let startNode = head;
  
  //Move previousNode to m and currentNode to m+1
  let flag;
  while(countm<m && currentNode !== null){
    if(!flag) flag = 1; //need to check whether m is whithin range
    //advance pointer and do nothing
    startNode = previousNode;
    previousNode = currentNode;
    currentNode = currentNode.next;
    countm++;
      
  }
  
  if(!flag) return head; //m doesn't exist within range
  
  //edge case when m=1
  if(startNode === previousNode) {
    while(countn < (n-m) && currentNode !== null){
  		//save original location
      temp = currentNode.next;

      //switch pointers
      currentNode.next = previousNode;
      startNode.next = temp;
      previousNode = currentNode;
      
      //advance node
      currentNode = temp;
      countn++;
    }
    return previousNode;
  }
  
  // when m>1
  while(flag=== 1 && countn < (n-m) && currentNode !== null){    
    //save original locations for advancement after switching pointers
    temp = currentNode.next;
    temp2 = previousNode;
    
    //switch pointers
    currentNode.next = startNode.next;
    startNode.next = currentNode;
    previousNode.next = temp;
    
    //advance prev & cur using the original locations
    previousNode = temp2;
    currentNode = temp;
    countn++;
  }
  
  return head;
};