技術標籤：leetcode#連結串列指標連結串列單鏈表演算法leetcode

迭代好寫多了，3個指標即可，最後當cur走到空時，pre就是新連結串列的頭

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head)
 
 {
        //先迭代，三個指標解決
        //從頭走到尾，記錄當前指標，前驅指標和後繼指標
        /*if(head == NULL || head->next == NULL) return head;
        ListNode* pre,*cur,*hou;
        pre = head;
        cur = head->next;
        hou = cur;
        pre->next = NULL;
        //當cur走到連結串列結束（null）時，pre就是新連結串列的頭部
        while(cur != NULL){
            hou = cur->next;
            cur->next = pre;
            pre = cur;
            cur = hou;
        }
        return pre;*/
 


        //遞迴
        if(head == NULL || head->next == NULL) return head;
        //在當前這一步反轉自己，讓自己的後繼結點指向自己
        ListNode* next = head->next;
        head->next = NULL;
        return helpReverse(head,next);
    }
    ListNode* helpReverse(ListNode* cur,ListNode* node){
        if(node == NULL
 
) return cur;
        ListNode* next = node->next;
        node->next = cur;
        return helpReverse(node,next);
    }
};