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  • LeetCode——重排链表

    Q:将给定的单链表L: L 0→L 1→…→L n-1→L n,重新排序为: L 0→L n →L 1→L n-1→L 2→L n-2→…
    要求使用原地算法,并且不改变节点的值
    例如:对于给定的单链表{1,2,3,4},将其重新排序为{1,4,2,3}.
    A:
    链表从中点分割成两个,后面的倒装后,再和前面的merge

        public static void reorderList(ListNode head) {
            if (head == null || head.next == null)
                return;
            ListNode slow = head;
            ListNode fast = head.next;
            while (fast != null && fast.next != null) {
                slow = slow.next;
                fast = fast.next.next;
            }
            fast = slow.next;
            slow.next = null;
            ListNode mid = reverse(fast);
            merge(head, mid);
        }
    
        public static ListNode reverse(ListNode head) {
            if (head == null || head.next == null)
                return head;
            ListNode head0 = new ListNode(Integer.MIN_VALUE);
            ListNode node = head;
            ListNode node1;
            while (node != null) {
                node1 = node.next;
                node.next = head0.next;
                head0.next = node;
                node = node1;
            }
            return head0.next;
        }
    
        public static void merge(ListNode head, ListNode mid) {
            ListNode node = head;
            ListNode node1 = mid;
            ListNode node2;
            while(node!=null && node1!=null){
                node2 = node1.next;
                node1.next = node.next;
                node.next = node1;
                node = node1.next;
                node1 = node2;
            }
        }
    
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  • 原文地址:https://www.cnblogs.com/xym4869/p/12443261.html
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