Prove safety offerNo15. Reverse list (Java)

Subject description:

After entering a list inverted list, the new list of the output header.

Thinking: with two pointers, one pointer is used to save the next node after head.next,

Then prev pointer to the previous node of the current node, the current node is a next node point to the front.

Finally, after the current node and the current node before a shift is performed

Code:

package offer;

public class TestNo15 {
    static class ListNode{
        int val;
        ListNode next;
        ListNode(int x){
            val = x;
        }
        public String toString(){
            if(this.next==null){
                return String.valueOf(this.val);
            }else{
                return this.val +"->" + this.next.toString();
            }
        }
    }
    public static void main(String[] args) {
        ListNode head = new ListNode(0);
        ListNode newHead = head;
        for(int i= 1;i<5;i++){
            newHead.next = new ListNode(i);
            newHead = newHead.next;
        }
        System.out.println(new TestNo15().ReverseList(head));
    }
    //非递归反转链表
    public ListNode ReverseList(ListNode head) {
        ListNode prev = null;
        ListNode next = null;
        while (head != null) {
            next = head.next; // next保存head的下一个节点
            head.next = prev; // 将当前节点的next断开，指向前一个节点，实现反转
            prev = head;//将当前节点的前节点进行后移
            head = next;//将当前节点进行后移
        }
        return prev;
    }
}