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;
}