I saw this problem in Leet and wanted to try it. one of the solutions proposed was the one below, however, I am scratching my head a little bit. I do understand what's going on but I don't understand some behaviors.

var reverseList = function(head){
  
  var tmp = null;
  var newHead = null;
  
  while(head !== null){
    tmp = head;
    head = head.next; 
    tmp.next = newHead; 
    newHead = tmp; 
  }
  
  return newHead;
}

For the input of [5,4,3,2,1] we get this in each iteration

[1,2,3,4,5]
[2,3,4,5]
null
[1]

[2,3,4,5]
[3,4,5]
[1]
[2,1]

[3,4,5]
[4,5]
[2,1]
[3,2,1]

[4,5]
[5]
[3,2,1]
[4,3,2,1]

[5]
null
[4,3,2,1]
[5,4,3,2,1]

here what's I am confused about

1. why is tmp = head; //tmp = [1,2,3,4,5] but when newHead = tmp // newHead = [1] and not [1,2,3,4,5]

2. head = head.next; doesn't seem to affect the two lines after it, but if we move head = head.next; to the last position, the code does not work

CodePudding user response：

why is tmp = head; //tmp = [1,2,3,4,5] but when newHead = tmp // newHead = [1] and not [1,2,3,4,5]

That is because between those two assignments, tmp.next has been reassigned. In the first iteration, tmp.next will be assigned null, so that effectively cuts the list short to just the node tmp itself.

head = head.next; doesn't seem to affect the two lines after it, but if we move head = head.next; to the last position, the code does not work

That's because it has to happen before tmp.next is assigned. If head = head.next has not happened before tmp.next = newHead is done, then tmp and head will still reference the same node, and thus tmp.next = newHead will mutate that same node. Hence, after that has happened head.next is no longer what it was before that assignment. This explains why head = head.next would not have the same effect anymore (and not the correct effect).