算法通关村|黄金挑战|K个一组进行反转

K个一组进行反转

1.头插法

public ListNode reverseKGroup(ListNode head, int k) {ListNode dummyNode = new ListNode(0);dummyNode.next = head;ListNode cur = head;// 计算链表长度int len = 0;while (cur != null) {len++;cur = cur.next;}// 计算有几组int n = len / k;ListNode pre = dummyNode;cur = head;// 一共进行n次反转for (int i = 0; i < n; i++) {// 每次反转操作将cur移动k-1次for (int j = 0; j < k - 1; j++) {ListNode next = cur.next;cur.next = cur.next.next;next.next = pre.next;pre.next = next;}pre = cur;cur = cur.next;}return dummyNode.next;
}

2.穿针引线法

public ListNode reverseKGroup(ListNode head, int k) {ListNode dummyNode = new ListNode(0);dummyNode.next = head;ListNode pre = dummyNode;ListNode end = dummyNode;while (end.next != null) {// 找到每次处理的一段区间的末尾处，移动k次for (int i = 0; i < k && end != null; i++) {end = end.next;}if (end == null) {break;}ListNode start = pre.next;ListNode next = end.next;// 与后边的链表断开end.next = null;// 调用反转方法，将断开的这一段链表反转，并且和前边的链表相连接pre.next = reverse(start);// 现在start是这段链表的末尾处，将其与后边的链表相连接start.next = next;// 把pre和end指针移动到下一个区间的前一个位置pre = start;end = pre;}return dummyNode.next;
}// 反转实现
private ListNode reverse(ListNode head) {ListNode pre = null;ListNode curr = head;while (curr != null) {ListNode next = curr.next;curr.next = pre;pre = curr;curr = next;}return pre;
}