C/java/python-链表逆序

• 链表逆序问题描述
• 一、C语言链表逆序
• • 1、直接逆序
  • 2、头插法逆序
  • 3、打印函数
  • 4、主函数
  • 5、运行结果
• 二、java链表逆序
• • 1、直接逆序
  • 2、头插法逆序
  • 3、打印函数
  • 4、主函数
  • 5、运行结果
• 三、Python链表逆序
• • 1、代码
  • 2、运行结果

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
解题思路：直接逆转链表 ：将设置两个变量，next保存当前的下一个节点，和一个中转的节点prev
第一步：将当前节点的下一个节点保存在next变量中；
第二步：将当前节点的下一个指向prev；
第三步：将head赋给prev，即prev向前移动；
第四步：将next赋值给head，即head向后移动；
接着循环，直到遍历完链表为止；

List链表的结构体:

struct List{
	int val;
	struct List *next;
};

// 直接逆转链表
struct List* dirReverseList(struct List* head){

	struct List *prev = NULL;
	struct List *next = NULL; 
	while(head){
		struct List *next = head->next;
	    //next = head->next;
		head->next = prev;
		prev = head;
		head = next;
	}
	return prev;
}

// 头插法链表逆转
struct List* reverseList(struct List* head){

	struct List temp;
	temp.next = NULL;
	while(head){
		struct List* next = head->next;
		head->next = temp.next;
		temp.next = head;
		head = next;
	}
	return temp.next;
}

void printList(List* head,const char* name){
	printf("%s:t",name);

	if(!head){
		printf("NULLn");
		return;
	}
	while(head){
		printf("[%d]", head->val);
		head = head->next;
	}
	printf("n");
}

	List a,b,c,d;
	a.val = 1;
	b.val = 2;
	c.val = 3;
	d.val = 4;
	a.next = &b;
	b.next = &c;
	c.next = &d;
	d.next = NULL;
	List* head = &a;
	printf("------------头插法/直接链表逆转-----------------n");
	printList(head,"old list");
	head =dirReverseList(&a);
	//head = reverseList(&a);
	printList(head,"new list");

old list:       [1][2][3][4]
new list:       [4][3][2][1]
Press any key to continue

// 迭代法，直接逆转
    public static ListNode iterator(ListNode head){
        ListNode temp=null,next;
        while(head!=null){
            next=head.next;
            head.next=temp;
            temp=head;
            head=next;
        }
        return temp;
    }

// 头插法
    public static ListNode headInsert(ListNode head){
        ListNode next;

        ListNode temp = new ListNode();

        temp.next=null;
        while (head != null){
            next = head.next;
            head.next = temp.next;
            temp.next = head;
            head = next;
        }
        return temp.next;
    }

    public static void printList(ListNode node){
        while (node != null){
            System.out.printf(node.val + " ");
            node = node.next;
        }
        System.out.println();
    }

        ListNode node6 = new ListNode(6,null);
        ListNode node5 = new ListNode(5,node6);
        ListNode node4 = new ListNode(4,node5);
        ListNode node3 = new ListNode(3,node4);
        ListNode node2 = new ListNode(2,node3);
        ListNode node1 = new ListNode(1,node2);

        printList(node1);
        ListNode reverseNode = iterator(node1);
        printList(reverseNode);
        //System.out.println("------------------");
        //ListNode node = headInsert(node1);
        //printList(node);

1 2 3 4 5 6 
------------------
6 5 4 3 2 1 

# coding: utf-8
# __author__="星仔"
# Email: 6316@qq.com
# Date: 2021/9/28 0028
class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


class Solution(object):
    @staticmethod
    def reverse(Head):
        temp = ListNode()
        while Head is not None:
            next = Head.next
            Head.next = temp.next
            temp.next = Head
            Head = next
        return temp.next


def printList(Head, name):
    print("%s:" % name, end="t")
    if Head is None:
        print('NULL')
        return
    while Head is not None:
        print('[%d]' % Head.val, end='')
        Head = Head.next
    print('')


if __name__ == '__main__':
    a = ListNode(1)
    b = ListNode(2)
    c = ListNode(3)
    d = ListNode(4)
    e = ListNode(5)

    a.next = b
    b.next = c
    c.next = d
    d.next = e

    printList(a, "old list")
    solution = Solution()
    head = solution.reverse(a)
    printList(head, 'reverse list')

E:pythonProjectvenvscriptspython.exe E:/pythonProject/9-28/algorithm/reverseList.py
old list:	[1][2][3][4][5]
reverse list:	[5][4][3][2][1]

进程已结束，退出代码为 0