标签:length 剑指offer system focus nod int == exp expand
题目描述
输入一个链表,反转链表后,输出新链表的表头。
1 public static ListNode ReverseList(ListNode head) { 2 if(head==null||head.next==null){ 3 return head; 4 } 5 Stack<ListNode> stack = new Stack<>(); 6 ListNode current=head; 7 while (current!=null){ 8 stack.push(current); 9 current=current.next; 10 } 11 //关键在于这里,原来的头结点的next要置为空,否则导致遍历时无限循环 12 head.next = null; 13 ListNode backHead=stack.pop(); 14 ListNode backCurrent=backHead; 15 while (!stack.isEmpty()){ 16 backCurrent.next=stack.pop(); 17 backCurrent=backCurrent.next; 18 } 19 return backHead; 20 }
1 //依次遍历所有节点,将所有节点的next指向前一个节点 2 public static ListNode ReverseList02(ListNode head) { 3 ListNode pre = null; 4 ListNode next =null; 5 while (head != null) { 6 //持有下一个节点的引用 7 next = head.next; 8 //将当前节点对下一个节点的引用指向前一个节点 9 head.next = pre; 10 //将前一个节点指向当前节点 11 pre = head; 12 //将当前节点指向下一个节点 13 head = next; 14 } 15 //如果head为null的时候,pre就为最后一个节点了,但是链表已经反转完毕,pre就是反转后链表的第一个节点 16 return pre; 17 }
1 public static ListNode ReverseList01(ListNode head) { 2 if(head == null || head.next == null) { 3 return head; 4 } 5 //递归找到尾部节点 6 ListNode preNode = ReverseList01(head.next); 7 //倒序 8 head.next.next = head; 9 //将正序的头节点指向空 10 head.next = null; 11 return preNode; 12 }
初始化链表:
1 public static class ListNode{ 2 int val; 3 ListNode next = null; 4 ListNode(int val) { 5 this.val = val; 6 } 7 } 8 public static ListNode createList(int[] list){ 9 ListNode head = new ListNode(-1); 10 ListNode current=head; 11 for(int i=0;i<list.length;i++){ 12 current.next=new ListNode(list[i]); 13 current=current.next; 14 } 15 return head.next; 16 }
测试:
1 public static void main(String[] args) { 2 int[] list={1,2,3,4,5,6}; 3 ListNode head = createList(list); 4 ListNode backHead = ReverseList(head); 5 while (backHead!=null){ 6 System.out.print(backHead.val+" "); 7 backHead=backHead.next; 8 } 9 } 10 输出:6 5 4 3 2 1
标签:length 剑指offer system focus nod int == exp expand
原文地址:https://www.cnblogs.com/Blog-cpc/p/12393117.html