25. K 个一组翻转链表

25. K 个一组翻转链表

思路

遍历链表–>如果每少于k个–>翻转操作
–>如果每到k个–>递归

总结

其实常规思路是 每k个进行一个翻转，然后把所有的翻转结果进行一个拼接，但是效率较低。用递归的思路则很不错。

具体实现

	public ListNode reverseKGroup(ListNode head, int k) {
        return swapPairsPerIntNum(head, k);
    }

    private ListNode swapPairsPerIntNum(ListNode head, int n) {

        if(n == 1){
            return head;
        }

        ListNode temp = new ListNode(0,head);

        ListNode result = new ListNode(), oldNext,tail = null,current;

        int index = 0;
        while(temp.next != null){
            index++;
            current = new ListNode(temp.next.val);

            if(index == 1){
                tail = current;
            }

            if(index <= n){
                oldNext = result.next;
                result.next = current;
                result.next.next = oldNext;

                temp = temp.next;
            }else{

                boolean hasNext = this.checkNum(temp.next, n);
                if(hasNext){
                    tail.next = swapPairsPerIntNum(temp.next,n);
                }else{
                    tail.next = temp.next;
                }

                break;
                
            }
        }

        return result.next;
    }


    private boolean checkNum(ListNode node, int k){
        ListNode temp = new ListNode(-1,node);

        int index = 0;

        while(temp.next != null){
            index++;
            temp = temp.next;

            if(index > k){
                break;
            }
        }

        return index >= k;
    }