Day 03 翻转链表

1、翻转链表

• while循环

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

• 递归

  核心在递归翻转newHead向前承接的过程

  class Solution {
      public ListNode reverseList(ListNode head) {
  	if (head == null || head.next == null){
  		return head;
  	}
  	ListNode newHead = reverseList(head.next);
  	head.next.next = head;
  	head.next = null;
  	return newHead;
    }
  }
  

2、翻转前k个链表

• 递归。第1题的扩展

  需要将head的next指向第k+1个链表

    func reverseK(head *ListNode, k int) *ListNode {
          if head.Next == nil || k == 1 {
                  return head
          }
          last := reverse_K(head.Next, k-1)
          temp := head.Next
          head.Next = temp.Next
          temp.Next = head
          return last
    }
  

3、区间翻转

• 前k个翻转的扩展

  当left为0，即为翻转前k个

    func reverseBetween(head *ListNode, left int, right int) *ListNode {
          if left == 1 {
                  return reverse_K(head, right)
          }
          head.Next = reverseBetween(head.Next,left - 1,right - 1)
          return head
    }
      func reverse_K(head *ListNode, k int) *ListNode {
          if head.Next == nil || k == 1 {
                  return head
          }
          last := reverse_K(head.Next, k-1)
          temp := head.Next
          head.Next = temp.Next
          temp.Next = head
          return last
    }
  

• for循环解法

  class Solution {
  /**
   * 1、找到左侧节点
   * 2、对区间内的节点进行翻转，记录区间外的节点头
   * 3、连接1 2 3
   * @param head 头结点
   * @param left 左侧
   * @param right 右侧
   * @return 翻转结果
   */
  public ListNode reverseBetween(ListNode head, int left, int right) {
  	if (head == null || left >= right ){
  		return head;
  	}
  	ListNode node = new ListNode(-1);
  	node.next = head;
  	head = node;
  	for (int i = 1; i < left; i++) {
  		head = head.next;
  	}
  	ListNode pre = head;
  	ListNode curr = head.next;
  	ListNode tail = curr;
  	ListNode next = null;
  	ListNode tempPre = tail;
  
  	for (int i = left; i <= right; i++) {
  		next = curr.next; //345  45  5
  		curr.next = tempPre;
  		tempPre = curr;//32
  		curr = next;//45
  	}
  	pre.next = tempPre;//1  432
  	tail.next = next;//5
  
  	return node.next;
  }
  

} ```