LEETCODE - Linked List 题目思路汇总

浏览了一下 Leetcode 上 Linked List 的题目，我把它分为 6 类：

调换顺序
删除
合并
环
变身
复制

做Leetcode还是要归类总结才好玩，最开始做两三个觉得很懵，做四五个就能发现规律，找到适合自己的思考方式，剩下的题就都迎刃而解。打通任督二脉后，做题也会上瘾，练练脑子还挺好玩的。

每个题的源码和详细思路可以看这里。

目前做完：

调换顺序：234，147，206，24，143，328，61
删除：19，83，203，82
合并：21，160
环：141，142
变身：109，2

今天周六，是玩的时间，不过晚上回来还是做了几个题，其他题目如果感兴趣的话可以去上面的链接里看。

[1] 调换顺序

328，Odd Even Linked List

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None: 
            return head
            
        odd = oddHead = head
        even = evenHead = head.next
        
        while even and even.next:
            odd.next = odd.next.next
            even.next = even.next.next
            odd = odd.next
            even = even.next
        odd.next = evenHead
        
        return oddHead

61，Rotate List

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def rotateRight(self, head, k):
        
        if head is None or head.next is None or k == 0:
            return head

        dummy = ListNode(0)
        dummy.next = head       


        # 计算length，求余数，求step
        p = dummy           #p从dummy开始，步数就可以和length保持一致
        length = 0
        while p.next:       #p最后是走到尾部，而不是None，这样可以连接指向原head
            length += 1
            p = p.next
        
        # 计算长度的同时，顺便把尾部指向原head
        p.next = head
        
        # 找到新head
        step = length - k % length

        while step > 0:     #p从尾部开始，相当于从dummy开始，步数和step保持一致 
            p = p.next
            step -= 1
            
        dummy.next = p.next
        p.next = None
        
        return dummy.next

[2] 删除

82，Remove Duplicates from Sorted List II

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteDuplicates(self, head):
        
        if head is None or head.next is None:
            return head
        
        dummy = ListNode(0)
        dummy.next = head
        
        last = dummy
        cur = head

        
        while cur and cur.next:
            
            p = cur.next
            
            if p.val != cur.val:
                last.next = cur
                last = cur
                cur = p

            else:
                while p and p.val == cur.val:  
                    p = p.next
                cur = p

                
        last.next = cur
        
        return dummy.next