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;

}