CCI_chapter 2 Linked Lists
2.1 Write code to remove duplicates from an unsorted linked list
/* Link list node */ struct node { int data; struct node* next; }; void rem_duplicate(node *head){ if(NULL == head) return ; set<int> hash; set.insert(head->data); while(head->next){ if(hash.find(head->next->data) == hash.end()){ node *tp = head->next; head->next = tp->next; delete tp; }else{ hash.insert(head->next->data); head = head ->next; } } }
2.2 Implement an algorithm to fnd the nth to last element of a singly linked list
/* Link list node */ struct node { int data; struct node* next; }; node *nthToLast(node * head, int n){ if(NULL == head || n <1) return NULL; node *p; p = head for(int i = 1; i < n; i++){ if(p== NULL) return NULL; p = p->next; } node * q = head; while(p->next){ p = p->next; q = q->next; } return p; }
2.3 Implement an algorithm to delete a node in the middle of a single linked list, given only access to that node
/*
Initialize mid element as head and initialize a counter as 0. Traverse the list from head, while traversing increment the counter and change mid to mid->next whenever the counter is odd. So the mid will move only half of the total length of the list.
*/
/* Link list node */
struct node
{
int data;
struct node* next;
};
void deleteMiddle(struct node *head){
if(head == NULL) return ;
node * mid = head;
int count = 0;
while(head != NULL){
if(count & 1){
mid = mid->next;
}
count++;
head = head ->next;
}
if(mid->next !=NULL)
{
mid->data = mid->next->data;
node *tp = mid->next;
mid->next = tp->next;
delete tp;
}else{
delete mid;
}
}
reference :http://www.geeksforgeeks.org/write-a-c-function-to-print-the-middle-of-the-linked-list/
2.4 You have two numbers represented by a linked list, where each node contains a single digit The digits are stored in reverse order, such that the 1’s digit is at the head of the list Write a function that adds the two numbers and returns the sum as a linked list
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { // Start typing your C/C++ solution below // DO NOT write int main() function if(l1 == NULL) return l2; if(l2 == NULL) return l1; int last = 0; ListNode *p, *q, *pre; p = l1; q= l2; ListNode *head = p; pre = NULL; while(p && q){ p ->val += q->val + last; if(p->val >= 10){ last = 1; p->val -= 10; }else{ last = 0; } pre = p; p = p->next; q = q->next; } p = NULL == q ? p : q; pre ->next = p; while(p && last == 1){ p->val += last; if(p->val >= 10){ last = 1; p->val -= 10; }else{ last = 0; } pre = p; p = p->next; } if(last == 1){ q = new ListNode(1); pre ->next = q; } return head ; } };
2.5 Given a circular linked list, implement an algorithm which returns node at the beginning of the loop
DEFINITION
Circular linked list: A (corrupt) linked list in which a node’s next pointer points to an
earlier node, so as to make a loop in the linked list
EXAMPLE
Input: A -> B -> C -> D -> E -> C [the same C as earlier]
Output: C
分析：最简洁易懂的还是数学形式的表达。设有两个指针p1、p2, p1每次向前移动一下，p2每次向前移动两个。p1到达loop 入口时，p2比p1多走了K个节点（k即为从链表入口到loop 的距离），假设T时刻两个节点相遇，则2T - T = n - k 即 T = n - K ,即相遇点到loop的入口距离为K。那么相遇点到Loop 的距离和链表头到loop 的距离相等，loop的入口点可求。
/* Link list node */ struct node { int data; struct node* next; }; // 假设输入的链表一定有环 node * FindBeginning(node * head){ if(head == NULL) return NULL; node *p, *q; p = head ; q = head; do{ p = p->next; q = q->next; q = q->next; }while(p != q) q = head ; while(q != p){ p = p->next; q = q->next ; } return p; }