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要解决链表的回文结构:首先需要求中间节点,其次是会反转链表。
typedef struct ListNode ListNode; struct ListNode* middleNode(struct ListNode* head) { ListNode* cur = head; int listLength = 0; while(cur) { //求链表的长度 listLength++; cur = cur->next; } //链表中间节点的位置 int middle = listLength / 2 + 1; int i = 1; //注意:非i=0 cur = head; while(i < middle) { i++; cur = cur->next; } return cur; }
两个
节点,slow指针一次走一个
节点,当fast != NULL && fast->next != NULL
时循环继续,否则循环结束。情况1.含有奇数
个节点
情况2.含有偶数
个节点
typedef struct ListNode ListNode;
struct ListNode* middleNode(struct ListNode* head)
{
//快慢指针:慢指针一次走一步,快指针一次走两步
ListNode* fast = head;
ListNode* slow = head;
//注意循环继续的条件是&&而不是||,且fast与fast->next的位置不能交换
while (fast != NULL && fast->next != NULL)
{
fast = fast->next->next;
slow = slow->next;
}
return slow;
}
length
。length - k
个节点。length - k
个节点,返回节点指针即可。typedef struct ListNode ListNode; int kthToLast(struct ListNode* head, int k) { //1.遍历链表求出链表长度,再遍历一次链表,找到返回值 int size = 0; ListNode* cur = head; while(cur) { size++; cur = cur->next; } int i = 0; cur = head; while(i < size - k) { cur = cur->next; i++; } return cur->val; }
k个节点
。fast != NULL
时循环继续,否则循环结束。typedef struct ListNode ListNode; int kthToLast(struct ListNode* head, int k) { //2.快慢指针:快指针先走k步,然后快指针一次走一步,慢指针一次走一步 ListNode* fast = head; ListNode* slow = head; for (int i = 0; i < k; i++) { fast = fast->next; } while (fast != NULL) { fast = fast->next; slow = slow->next; } return slow->val; }
typedef struct ListNode ListNode; struct ListNode* reverseList(struct ListNode* head) { //1.创建新链表,遍历原链表,逐个头插 ListNode* newHead = NULL, *cur = head; while(cur) { //头插 ListNode* next = cur->next; cur->next = newHead; newHead = cur; cur = next; } return newHead; }
typedef struct ListNode ListNode; struct ListNode* reverseList(struct ListNode* head) { //2.创建三个指针,反转指针的指向 if(head == NULL) { return NULL; } ListNode* n1 = NULL, *n2 = head, *n3 = n2->next; while(n2) { n2->next = n1; n1 = n2; n2 = n3; if(n3 != NULL) { n3 = n3->next; } } return n1; }
class PalindromeList { public: //判断数组是否满足回文结构 bool isReverse(int arr[], int left, int right) { while(left < right) { if(arr[left] != arr[right]) { return false; } left++; right--; } return true; } bool chkPalindrome(ListNode* A) { int arr[900]; ListNode* cur = A; int i = 0, listLength = 0; while(cur) { arr[i++] = cur->val;//将链表中的值保存到数组中 cur = cur->next; listLength++;//求链表的长度 } return isReverse(arr, 0, listLength - 1); } };
情况1.含有奇数
个节点:
情况2.含有偶数
个节点:
class PalindromeList { public: ListNode* findMidNode(ListNode* phead) { ListNode* fast = phead; ListNode* slow = phead; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; } return slow; } ListNode* reverseList(ListNode* phead) { ListNode* n1, *n2, *n3; n1 = NULL, n2 = phead, n3 = n2->next; while(n2) { n2->next = n1; n1 = n2; n2 = n3; if(n3 != NULL) { n3 = n3->next; } } return n1; } bool chkPalindrome(ListNode* A) { //1.找链表的中间节点 ListNode* mid = findMidNode(A); //2.反转中间节点以及之后的节点组成的链表 ListNode* right = reverseList(mid); //3.遍历反转链表,与原链表进制值的比较 ListNode* left = A; while(right) { if(right->val != left->val) { return false; } right = right->next; left = left->next; } return true; } };
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