no206 Reverse Linked List

最新推荐文章于 2022-10-17 16:52:17 发布

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本文深入探讨了链表逆序算法的实现细节，通过迭代和递归两种方法，详细解释了如何反转链表。提供了完整的C++代码示例，并对比了使用栈进行逆序的替代方案。

#include<iostream>

using namespace std;

struct ListNode{
	int val;
	ListNode *next;
	ListNode(int x):val(x),next(NULL){}
};
//https://leetcode-cn.com/problems/reverse-linked-list/solution/die-dai-di-gui-jie-fa-by-sunshy/
class Solution {
public:
	ListNode* reverseList(ListNode* head) {
       if(!head){
			return nullptr;
		} 
		ListNode *pre = head;
		ListNode *cur = head->next;
		while(cur !=nullptr){
			pre->next = cur->next;
			ListNode *tmp= cur->next;
			cur->next = head;
			head = cur;
			cur = tmp;
		}		
		return head;
	}
};
int main(){
	Solution so;
	ListNode *l1 = new ListNode(1);
	ListNode *l2 = new ListNode(2);
	ListNode *l3 = new ListNode(3);
	ListNode *l4 = new ListNode(4);
	ListNode *l5 = new ListNode(5);

	l1->next = l2;
	l2->next = l3;
	l3->next = l4;
	l4->next = l5;
	ListNode *ll = so.reverseList(l1);
	while(ll->next !=NULL){
		cout << l1->val << ",";	
		l1 = l1->next;
	}
}

//use stack
    public static ListNode reverseListVersion2(ListNode head) {
        if(head == null){
            return null;
        }
        Stack<Integer> transferStack = new Stack<>();
        ListNode pre =  head;
        while (pre.next!=null){
            transferStack.push(pre.val);
            pre = pre.next;
        }
        transferStack.push(pre.val);
        ListNode dummpy = new ListNode(-1);
        pre = dummpy;
        while (transferStack.size()>0){
            pre.next = new ListNode(transferStack.pop());
            //新链表每加一个节点，当前指针就向前走一步
            pre = pre.next;
        }return dummpy.next;
    }