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Reverse a singly linked list.
代码如下:
the iterative solution:(c++)
/**
* 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) {
ListNode *temp = NULL , *nextNode = NULL;
while(head){
nextNode = head->next; //save the current‘s next node
head->next = temp; //let the current point to its previous one
temp = head; //save the current node as pre
head = nextNode; //move to next node
}
return temp; //just point to the last node we wanted
}
};
或:
class Solution { public: ListNode *reverseList(ListNode *head) { if (head == NULL || head->next == NULL) return head; ListNode *pCurr = head; ListNode *pPrev = NULL; ListNode *pNext = NULL; while (pCurr != NULL) { pNext = pCurr->next; //save next node pCurr->next = pPrev; if (pNext == NULL) head = pCurr; pPrev = pCurr; pCurr = pNext; } return head; } };
其他解法:
1、 the recursive version:(c++)
class Solution { public: ListNode* reverseList(ListNode* head) { if (head == NULL || head->next == NULL) return head; // case proof and recursion exit conditon ListNode *np = reverseList(head->next); head->next->next = head; // make the next node point back to the node itself head->next = NULL; // destroy the former pointer return np; } };
2、(c++)
class Solution { public: ListNode* reverseList(ListNode* head) { stack<ListNode*> s; ListNode *tail=NULL; while(head) { s.push(head); head=head->next; } if(!s.empty()) head=s.top(); while(!s.empty()) { tail=s.top(); s.pop(); if(!s.empty()) tail->next=s.top(); else tail->next=NULL; } return head; } };
3、(c)
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* reverseList(struct ListNode* head) { struct ListNode* last = 0; while (head) { struct ListNode* next = head->next; head->next = last; last = head; head = next; }; return last; }
或:
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* reverseList(struct ListNode* head) { if (!head) return NULL; struct ListNode *curr = head; struct ListNode *next = NULL; struct ListNode *prev = NULL; while (curr){ next = curr->next; curr->next = prev; prev = curr; curr = next; head = prev; } return head; }
更多:http://blog.chinaunix.net/uid-7471615-id-83821.html
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原文地址:http://www.cnblogs.com/carsonzhu/p/4589839.html