algorithm - Reverse LinkedList - 链表工具

Question

这是 LeetCode 中的一个简单问题:反向链表我有两个相似的代码，我无法弄清楚两者之间的区别，但它输出不同的结果。

第一个 while 循环的结果是正确的答案，但第二个 while 循环的答案是错误的。而不是 temp = current。接下来，我只是将电流存储到温度在 while 循环的最后一行，我正确地将 current 切换为 temp.next。我认为他们应该得到相同的答案但是当输入 {1,2,3,4,5} 时，第二个解决方案得到了错误的答案 {1}。

ListNode reverse = null;
ListNode current = head;

while(current != null){
  ListNode temp = current.next;
  current.next = reverse;
  reverse = current;
  current = temp;
}

这是第二个 while 循环。

while(current != null){
  ListNode temp = current;
  current.next = reverse;
  reverse = current;
  current = temp.next;
}

Answer 1

while(current != null){
    // This line temp is keeping the reference  of current variable
    ListNode temp = current;
    // now temp and current both pointing to same value

    // This line will make the next of current variable pointing to reverse reference
    // as current and temp were same so next of temp is also pointing to the reverse
    current.next = reverse;

    // now the reverse will be changed and point to the current reference
    reverse = current;
    // so up to here reverse, temp, current all pointing to the same location
    // as all are pointing same so temp.next,curr.next and reverse.next
    // will store the previous reference 

    // current will point back to the reverse value
    // no increment is done in the whole loop
    current = temp.next;
}