Leetcode-25:Reverse Nodes in k-Group

2018-01-09

问题描述：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

问题示例

Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5

我的答案

//最初的想法:每k个节点调用一次reverse函数(单独写一个reverse函数，用回溯返回节点来倒置，但是总是有BUG，有机会重写)
//该版本在不足k个时，对k个进行了置换
ListNode* reverseKGroup(ListNode* head, int k) {
    ListNode *lefthead;
    ListNode *reversehead;
    ListNode *reverseend;
    ListNode *nextLoop;
    ListNode *temp;
    if(!head){
        return NULL;
    }
    reversehead = head;
    reverseend = head;
    lefthead = head->next;
    for(int i = k;i>0;i--){
        if(head->next){
            nextLoop = head->next;
            head = head->next;
        }else{
            nextLoop = NULL;
        }
    }
    int j = k;
    while (j>1&&lefthead) {
        j--;
        temp = lefthead->next;
        lefthead->next = reversehead;
        reversehead = lefthead;
        lefthead = temp;
    }
    if(j == 1&&nextLoop){
        reverseend->next = reverseKGroup(nextLoop,k);
    }else{
        reverseend->next = NULL;
    }
    return reversehead;
}
//AC
ListNode* reverseKGroup(ListNode* head, int k) {
    ListNode *lefthead;
    ListNode *reversehead;
    ListNode *reverseend;
    ListNode *nextLoop;
    ListNode *temp;
    ListNode *checkLeft;
    if(!head){
        return NULL;
    }else if(k==1){
        return head;
    }
    
    reversehead = head;
    reverseend = head;
    lefthead = head->next;
    for(int i = k;i>0;i--){
        if(head->next){
            nextLoop = head->next;
            head = head->next;
        }else{
            nextLoop = NULL;
        }
    }
    int leftnumber = 1;
    checkLeft = reversehead;
    while(checkLeft->next&&leftnumber<k){
        checkLeft = checkLeft->next;
        leftnumber++;
    }
    
    if(leftnumber==k){
        for(int i=k;i>1;i--){
            temp = lefthead->next;
            lefthead->next = reversehead;
            reversehead = lefthead;
            lefthead = temp;
        }
    }
    
    if(leftnumber==k){
        reverseend->next = reverseKGroup(nextLoop,k);
    }
    
    return reversehead;
}

其他高效答案

ListNode* reverseKGroup(ListNode* head, int k) {
        if (!head || !head->next) return head;
        ListNode* h = head;
        int len = 0;
        ListNode l(0);
        l.next = head;
        ListNode* pre = &l;
        
        while (h) {
            h = h->next;
            len++;
        }
        int n = len/k;
        h = head;
        for (int i = 0; i < n; i++) {
            for (int j = 1; j < k; j++) {
                ListNode* next = h->next;
                ListNode* nextnext = h->next->next;
                next->next = pre->next;
                pre->next = next;
                h->next = nextnext;
            }
            pre = h;
            h = h->next;
        }
        return l.next;
    }