链表题目---4 删除链表中重复的结点和判断链表是否为回文链表

删除链表中重复的结点

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/*
struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}
};
*/
class Solution {
public:ListNode* deleteDuplication(ListNode* pHead){if(pHead == NULL){return NULL;}ListNode *prev = NULL; //用于删除的结点, 是p1的前一个结点ListNode *p1 = pHead;ListNode *p2 = pHead->next;ListNode *result = pHead;  //用于返回的第一个结点while(p2 != NULL){if(p1->val != p2->val){prev = p1;p1 = p2;p2 = p2->next;}else{while(p2!=NULL && (p2->val == p1->val)){p2 = p2->next;}if(prev != NULL){prev->next = p2;}else{result = p2;}p1 = p2;if(p2!=NULL){p2 = p2->next;}}}return result;}
};

判断链表是否为回文链表

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思路

找到中间结点，如果是偶数，就找第二个
反转从中间结点后面的单链表
然后两个链表进行比较，每个值必须一样，长度不用管，因为如果是回文，长度在找中间结点的时候就已经定下了

/*
struct ListNode {int val;struct ListNode *next;ListNode(int x) : val(x), next(NULL) {}
};*/class PalindromeList {
public://找中间结点,如果是偶数找第二个ListNode* middleNode( ListNode* head){if(head == NULL){return NULL;}ListNode *fast =head;ListNode *slow =head;while(1){fast =fast->next;if(fast == NULL){break;}fast = fast->next;if(fast == NULL){break;}slow = slow->next;}return slow;}//反转单链表ListNode *reverseList( ListNode* head){if(head == NULL){return head;}ListNode *prev,*cur,*next;prev = NULL;cur = head;next = head->next;while(cur != NULL){cur->next = prev;prev = cur;cur =next;if(next != NULL){next=next->next;}}return prev;}bool chkPalindrome(ListNode* A) {ListNode * middle = middleNode(A);ListNode * r = reverseList(middle);ListNode *n1 = A, *n2 =r;while(n1 !=NULL && n2 !=NULL){if(n1->val != n2->val){return false;}n1 = n1->next;n2 = n2->next;}return true;}
};