83. Remove Duplicates from Sorted List [easy] (Python)__Python

題目連結

https://leetcode.com/problems/remove-duplicates-from-sorted-list/ 題目原文

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3. 題目翻譯

給定一個有序鏈表，重複資料刪除元素使每個元素之出現一次。 思路方法 思路一

遍曆所有節點，對於每個節點，檢查其後的一個節點是否與當前節點值相同，若相同則刪除後面的節點。迴圈下去。

代碼

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def deleteDuplicates(self, head):        """        :type head: ListNode        :rtype: ListNode        """        p = head        while p:            if p.next and p.next.val == p.val:                p.next = p.next.next            else:                p = p.next        return head

思路二

遍曆所有節點，對於每個節點，從後一個節點開始一個個檢查是否與當前節點值相同，直到找到一個後面的節點其值與當前節點不同，刪除中間的所有與當前節點值相同的節點。迴圈下去。

代碼

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def deleteDuplicates(self, head):        """        :type head: ListNode        :rtype: ListNode        """        p = q = head        while p:            while q and q.val == p.val:                q = q.next            p.next = q            p = q        return head

思路三

遞迴實現，本質上是從後向前檢查是否有相同的節點。

代碼

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def deleteDuplicates(self, head):        """        :type head: ListNode        :rtype: ListNode        """        if not head or not head.next:            return head        head.next = self.deleteDuplicates(head.next)        return head if head.val != head.next.val else head.next

PS: 新手刷LeetCode，新手寫部落格，寫錯了或者寫的不清楚還請幫忙指出，謝謝。
轉載請註明：http://blog.csdn.net/coder_orz/article/details/51506143