回文链表 牛客网 程序员面试金典 C++ Python
回文链表 牛客网 程序员面试金典 C++ Python
- 题目描述
- 请编写一个函数,检查链表是否为回文。
- 给定一个链表ListNode* pHead,请返回一个bool,代表链表是否为回文。
- 测试样例:
- {1,2,3,2,1}
- 返回:true
- {1,2,3,2,3}
- 返回:false
class Palindrome {
public:
// run:4ms memory:492k
bool isPalindrome(ListNode* pHead) {
if(NULL == pHead) return true;
if(NULL == pHead->next) return true;
if(NULL == pHead->next->next)
if(pHead->val == pHead->next->val) return true;
else return false;
bool ret = true;
ListNode* lastNode = NULL;
ListNode* nextNode = pHead->next;
ListNode* pSlow = pHead;
ListNode* pFast = pHead;
while(pFast && pFast->next){
pFast = pFast->next->next;
pSlow->next = lastNode;
lastNode = pSlow;
pSlow = nextNode;
nextNode = nextNode->next;
}
ListNode* lNode = lastNode;
ListNode* nNode = pSlow;
if (NULL == pFast)
if (pSlow->val != lastNode->val) ret = false;
else lastNode = lastNode->next;
while(lastNode){
if (lastNode->val != nextNode->val) {
ret = false;
break;
}
lastNode = lastNode->next;
nextNode = nextNode->next;
}
while(lNode){
ListNode* tmp = lNode->next;
lNode->next = nNode;
nNode = lNode;
lNode = tmp;
}
return ret;
}
// run:5ms memory:600k
bool isPalindrome2(ListNode* pHead) {
if(NULL == pHead) return true;
if(NULL == pHead->next) return true;
if(NULL == pHead->next->next)
if(pHead->val == pHead->next->val) return true;
else return false;
ListNode* lastNode = NULL;
ListNode* nextNode = pHead->next;
ListNode* pSlow = pHead;
ListNode* pFast = pHead;
while(pFast && pFast->next){
pFast = pFast->next->next;
pSlow->next = lastNode;
lastNode = pSlow;
pSlow = nextNode;
nextNode = nextNode->next;
}
if (NULL == pFast)
if (pSlow->val != lastNode->val) return false;
else lastNode = lastNode->next;
while(lastNode){
if (lastNode->val != nextNode->val) return false;
lastNode = lastNode->next;
nextNode = nextNode->next;
}
return true;
}
// run:4ms memory:476k
bool isPalindrome3(ListNode* pHead) {
if(NULL == pHead) return true;
if(NULL == pHead->next) return true;
if(NULL == pHead->next->next)
if(pHead->val == pHead->next->val)
return true;
else return false;
ListNode* lastNode = NULL;
ListNode* nextNode = pHead->next;
ListNode* pSlow = pHead;
ListNode* pFast = pHead;
while(pFast && pFast->next){
pFast = pFast->next->next;
pSlow->next = lastNode;
lastNode = pSlow;
pSlow = nextNode;
nextNode = nextNode->next;
}
if (NULL == pFast){
if (pSlow->val != lastNode->val)
return false;
else{
lastNode = lastNode->next;
while(lastNode){
if (lastNode->val != nextNode->val)
return false;
lastNode = lastNode->next;
nextNode = nextNode->next;
}
}
}
if(NULL == pFast->next){
while(lastNode){
if (lastNode->val != nextNode->val)
return false;
lastNode = lastNode->next;
nextNode = nextNode->next;
}
}
return true;
}
};
Python
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Palindrome:
# run:43ms memory:5732k
def isPalindrome(self, pHead):
if None == pHead: return True
if None == pHead.next: return True
if None == pHead.next.next:
if pHead.val == pHead.next.val: return True
else: return False
ret = True
lastNode = None
nextNode = pHead
pFast = pHead
pSlow = pHead
while pFast and pFast.next:
pFast = pFast.next.next
nextNode = pSlow.next
pSlow.next = lastNode
lastNode = pSlow
pSlow = nextNode
nextNode = nextNode.next
lNode = lastNode
nNode = pSlow
if None == pFast:
if pSlow.val != lastNode.val:ret = False
else:lastNode = lastNode.next
while lastNode and nextNode:
if lastNode.val != nextNode.val:
ret = False
lastNode = lastNode.next
nextNode = nextNode.next
while lNode:
tmp = lNode.next
lNode.next = nNode
nNode = lNode
lNode = tmp
return ret
# run:34ms memory:5728k
def isPalindrome2(self, pHead):
if None == pHead: return True
if None == pHead.next: return True
if None == pHead.next.next:
if pHead.val == pHead.next.val: return True
else: return False
lastNode = None
nextNode = pHead
pFast = pHead
pSlow = pHead
while pFast and pFast.next:
pFast = pFast.next.next
nextNode = pSlow.next
pSlow.next = lastNode
lastNode = pSlow
pSlow = nextNode
nextNode = nextNode.next
if None == pFast:
if pSlow.val != lastNode.val:return False
else:lastNode = lastNode.next
while lastNode:
if lastNode.val != nextNode.val:return False
lastNode = lastNode.next
nextNode = nextNode.next
return True