63. Swap nodes in pairs & rotate list & remove nth node from end of list

Swap nodes in pairs

Given a linked list, swap every two adjacent nodes and return its head.

For example, given1->2->3->4, You shoshould return the list2->1->4->3.

Your algorithm shocould use only constant space. You mayNotModify the values in the list, only nodes itself can be changed.

Note: during the first two SWAps, the head must change its position. There is also a pre pointer.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *swapPairs(ListNode *head) {        ListNode *pre = NULL, *p = head;        while(p && p->next) {            ListNode *q = p->next;            p->next = q->next;            q->next = p;            if(pre == NULL) head = q;            else pre->next = q;            pre = p;            p = p->next;        }        return head;    }};

 

Rotate list

Given a list, rotate the list to the rightKPlaces, whereKIs non-negative.

For example: Given1->2->3->4->5->NULLAndK=2, Return4->5->1->2->3->NULL.

Note: during the first two SWAps, the head must change its position. There is also a pre pointer.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */int getLength(ListNode *head) {    int len = 0;    while(head) {        ++len;        head = head->next;    }    return len;}class Solution {public:    ListNode *rotateRight(ListNode *head, int k) {        if(head == NULL) return NULL;        k = k % getLength(head);        if(k == 0) return head;        ListNode *first, *second, *preFirst;        first = second = head;        for(int i = 1; i < k && second->next; ++i) // k-1 step            second = second->next;        //if(second->next == NULL) return head;        while(second->next) {            preFirst = first;            first = first->next;            second = second->next;        }        second->next = head;        preFirst->next = NULL;        return first;    }};

 

Remove nth node from end of list

Given a linked list, removeNTh node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:GivenNWill always be valid. Try to do this in one pass.

Idea: Double pointer.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *removeNthFromEnd(ListNode *head, int n) {        ListNode *pre = NULL, *p1, *p2;        p1 = p2 = head;        for(int i = 1; i < n; ++i) p2 = p2->next;        while(p2->next) {            pre = p1;            p1 = p1->next;            p2 = p2->next;        }        if(pre) pre->next = pre->next->next;        return pre ? head : head->next;            }};

 

63. Swap nodes in pairs & rotate list & remove nth node from end of list