Another list manipulation problem.
class Solution { public: ListNode *reverseKGroup(ListNode *head, int k) { if (!head) return head; if (!head->next) return head; if (k == 1) return head; ListNode dum(-1); dum.next = head; ListNode *pProb = &dum; while (pProb) { // Prob ListNode *pPre = pProb; int cnt = k; while(cnt-- && pProb) pProb = pProb->next; // Valid? (= k) if(pProb) { ListNode *pEnd = pProb->next; ListNode *pToConn = pPre; ListNode *pOrig = pPre->next; for(int i = k; i >= 1; i --) { ListNode *p = pToConn; int cnt0 = i; while(cnt0--) p = p->next; pToConn->next = p; pToConn = p; p->next = pOrig; } pProb = pToConn; pToConn->next = pEnd; } } return dum.next; } };
LeetCode "Reverse Nodes in k-Group",布布扣,bubuko.com
LeetCode "Reverse Nodes in k-Group"
原文:http://www.cnblogs.com/tonix/p/3893605.html