題目描述

在一個排序的連結串列中，存在重複的結點，請刪除該連結串列中重複的結點，重複的結點不保留，返回連結串列頭指標。 例如，連結串列1->2->3->3->4->4->5 處理後為 1->2->5

解題思路

自己寫的程式碼太冗長，實在難看；牛客上找到的思路用Python實現的過程中，第一眼覺得挺簡單；後來越看越覺得好像其中連結串列的運用有些溜，反覆琢磨了好幾遍，難道是晚上思路卡頓了？

def deleteDuplication(self, pHead):
    first = ListNode(-1)
    first.next = pHead
    last = first
    while
 
 pHead and pHead.next:
        if pHead.val == pHead.next.val:
            val = pHead.val
            while pHead and val==pHead.val:
                pHead = pHead.next
            last.next = pHead
        else:
            last = pHead
            pHead = pHead.next
    return first.next

包含測試版：

class
 
 Solution:
    def deleteDuplication(self, pHead):
        first = ListNode(-1)
        first.next = pHead
        last = first
        while pHead and pHead.next:
            if pHead.val == pHead.next.val:
                val = pHead.val
                while pHead and val==pHead.val:
                    pHead = pHead.next
                last.next = pHead
            else
 
:
                last = pHead
                pHead = pHead.next
        return first.next

    def getNewChart(self, list):
        if list:
            node = ListNode(list.pop(0))
            node.next = self.getNewChart(list)
            return node

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


if __name__ == '__main__':
    list = [1,1,1,2,3,3,4]
    listNode = Solution().getNewChart(list)
    head = Solution().deleteDuplication(listNode)
    while head:
        print(head.val, end=" ")
        head = head.next