126 lines
2.4 KiB
C++
126 lines
2.4 KiB
C++
// Source : https://oj.leetcode.com/problems/reverse-linked-list-ii/
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// Author : Hao Chen
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// Date : 2014-07-05
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/**********************************************************************************
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*
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* Reverse a linked list from position m to n. Do it in-place and in one-pass.
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*
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* For example:
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* Given 1->2->3->4->5->NULL, m = 2 and n = 4,
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*
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* return 1->4->3->2->5->NULL.
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*
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* Note:
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* Given m, n satisfy the following condition:
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* 1 ≤ m ≤ n ≤ length of list.
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*
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*
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**********************************************************************************/
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#include <stdio.h>
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#include <stdlib.h>
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#include <time.h>
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struct ListNode {
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int val;
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ListNode *next;
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ListNode(int x) : val(x), next(NULL) {}
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};
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ListNode *reverseBetween(ListNode *head, int m, int n) {
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if (head==NULL || m>=n) return head;
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ListNode fake(0);
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fake.next = head;
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ListNode *pBegin=&fake, *pEnd=&fake;
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int distance = n - m ;
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while(pEnd && distance>0){
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pEnd = pEnd->next;
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distance--;
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}
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while(pBegin && pEnd && m-1>0) {
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pBegin = pBegin->next;
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pEnd = pEnd->next;
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m--;
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}
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if (pBegin==NULL || pEnd==NULL || pEnd->next == NULL){
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return head;
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}
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ListNode *p = pBegin->next;
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ListNode *q = pEnd->next->next;
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ListNode *pHead = q;
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while(p != q){
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ListNode* node = p->next;
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p->next = pHead;
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pHead = p;
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p = node;
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}
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pBegin->next = pHead;
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return fake.next;
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}
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void printList(ListNode* h)
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{
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while(h!=NULL){
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printf("%d ", h->val);
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h = h->next;
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}
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printf("\n");
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}
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ListNode* createList(int *a, int n)
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{
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ListNode *head=NULL, *p=NULL;
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for(int i=0; i<n; i++){
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if (head == NULL){
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head = p = new ListNode(a[i]);
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}else{
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p->next = new ListNode(a[i]);
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p = p->next;
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}
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}
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return head;
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}
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ListNode* createList(int len) {
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int *a = new int[len];
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for(int i=0; i<len; i++){
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a[i] = i+1;
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}
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ListNode* h = createList(a, len);
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delete[] a;
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return h;
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}
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int main(int argc, char** argv)
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{
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int l=5;
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int m=2, n=4;
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if (argc>1){
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l = atoi(argv[1]);
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}
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if (argc>2) {
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m = atoi(argv[2]);
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}
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if (argc>3) {
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n = atoi(argv[3]);
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}
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ListNode* h = createList(l);
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printList( h );
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printList( reverseBetween(h , m, n) );
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}
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