一个技巧秒杀翻转链表

文档阅读

文档阅读I
文档阅读II
模板
按照顺序拼接node，并可以递归处理

//翻转从[a,b)之间的元素 
    private ListNode reverse(ListNode a, ListNode b){
        ListNode pre = null, next = null;
        while(a != b){
            next = a.next;
            a.next = pre;
            pre = a;
            a = next;
        }
        return pre;
    }

题目

206. 反转链表

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null, next = null;
        while(head != null){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}

92. 反转链表 II

class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        ListNode dumpy = new ListNode(-999);
        dumpy.next = head;
        ListNode help = dumpy;
        for(int i = 0; i < left - 1; i++){
            help = help.next;
        }
        ListNode leftNode = help;
        help = help.next;
        ListNode startNode = help;//要翻转的头节点
        for(int i = left; i <= right; i++){
            help = help.next;
        }
        ListNode rightNode = help;
        ListNode newHead = reverse(startNode, rightNode);
        //拼接成新的
        leftNode.next = newHead;
        startNode.next = rightNode;
        return dumpy.next;
    }
    
    //翻转从[a,b)之间的元素 
    public ListNode reverse(ListNode head, ListNode tail) {
        ListNode pre = null, next = null;
        while(head != tail){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}

25. K 个一组翻转链表

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if(head == null) return null;
        ListNode a = head, b = a;
        for(int i = 0; i < k; i++){
            if(b == null) return head;
            b = b.next;
        }
        ListNode newHead = reverse(a, b);
        a.next = reverseKGroup(b, k);
        return newHead;
    }
    //翻转从[a,b)之间的元素 
    private ListNode reverse(ListNode a, ListNode b){
        ListNode pre = null, next = null;
        while(a != b){
            next = a.next;
            a.next = pre;
            pre = a;
            a = next;
        }
        return pre;
    }
}