问题描述：
struct list_node { int data; list_node* next; };
分析：
1 // 13_1.cc 2 #include <iostream> 3 using namespace std; 4 5 struct list_node { 6 int data; 7 list_node* next; 8 }; 9 10 list_node* find_kth(list_node* head, size_t k) { 11 if (!head) 12 return NULL; 13 14 // 统计链表中节点的个数 15 size_t count = 1; 16 list_node* cur = head; 17 while(cur->next != NULL) { 18 cur = cur->next; 19 count++; 20 } 21 22 if(count < k) 23 return NULL; 24 25 cur = head; 26 for(size_t i = 1; i <= count - k; i++) 27 cur = cur->next; 28 29 return cur; 30 } 31 32 // 插入元素 33 void insert_node(list_node*& head, int data) { 34 list_node* p = new list_node; 35 p->data = data; 36 p->next = head; 37 head = p; 38 } 39 40 int main() { 41 // 创建链表 42 list_node* head = NULL; 43 for (int i = 10; i > 0; i--) 44 insert_node(head, i); 45 46 // 查找倒数第k个 47 list_node* p = find_kth(head, 4); 48 cout << p->data << endl; 49 return 0; 50 }
方法2：
1 // 13_2.cc 2 #include <iostream> 3 using namespace std; 4 5 struct list_node { 6 int data; 7 list_node* next; 8 }; 9 10 list_node* find_kth(list_node* head, size_t k) { 11 if (!head) 12 return NULL; 13 14 list_node* p1 = head; 15 for (size_t i = 1; i <= k; i++) { 16 if (!p1) // 链表长度不足k 17 return NULL; 18 else 19 p1 = p1->next; 20 } 21 22 list_node* p2 = head; 23 while (p1) { 24 p1 = p1->next; 25 p2 = p2->next; 26 } 27 return p2; 28 } 29 30 // 插入元素 31 void insert_node(list_node*& head, int data) { 32 list_node* p = new list_node; 33 p->data = data; 34 p->next = head; 35 head = p; 36 } 37 38 int main() { 39 // 创建链表 40 list_node* head = NULL; 41 for (int i = 10; i > 0; i--) 42 insert_node(head, i); 43 44 // 查找倒数第k个 45 list_node* p = find_kth(head, 4); 46 cout << p->data << endl; 47 return 0; 48 }
转载自源代码
IT公司100题-13-求链表中倒数第k个结点,布布扣,bubuko.com
原文：http://www.cnblogs.com/dracohan/p/3901575.html