python链表 —— 如何从无序链表中移除重复项

方法一： 顺序删除

算法思想：

通过双层循环直接在链表上删除。外层循环用一个指针从第一个节点开始遍历整个链表，然后内层循环用另一个指针遍历其余节点与外层循环指针所指的节点的数据进行对比，若是相同则删除

性能分析：

该算法的时间复杂度为o(n*2)，空间复杂度为o(n)

#方法一：顺序删除
class LNode(object):
    def __init__(self, item):
        self.data = item
        self.next = None

def removeDup(head):
    if head == None or head.next == None:
        return
    outerCur = head.next
    interCur = None
    interPre = None
    while outerCur != None:
        interCur = outerCur.next
        interPre = outerCur
        while interCur != None:
            if outerCur.data == interCur.data:
                interPre.next = interCur.next
                interCur = interPre.next
            else:
                interPre = interCur
                interCur = interCur.next
        outerCur = outerCur.next

if __name__ == "__main__":
    i = 1
    head = LNode(0)
    head.next = None
    tmp = None
    cur = head
    while i < 7:
        tmp = LNode(i)
        if i%2 == 0:
            tmp.data = i+1
        elif i%3 == 0:
            tmp.data = i-2
        else:
            tmp.data = i
        tmp.next = None
        cur.next = tmp
        cur = tmp
        i += 1
    print("删除重复节点前：")
    cur = head.next
    while cur != None:
        print(cur.data)
        cur = cur.next
    removeDup(head)
    print("删除节点后：")
    cur = head.next
    while cur != None:
        print(cur.data)
        cur = cur.next

代码执行结果：

删除重复节点前：
1
3
1
5
5
7
删除节点后：
1
3
5
7

 

方法二：空间换时间

算法思想：

1. 建立一个列表保存已经遍历过的节点的值

2. 从头开始遍历链表，如果节点内容不再列表中，那么继续遍历链表，若已经存在，则删除该节点

性能分析：

时间复杂度为o(N)，空间复杂度为o(N)，以空间换时间

class LNode(object):
    def __init__(self, item):
        self.data = item
        self.next = None



def removeDup2(head):
    if head == None or head.next == None:
        return
    cur = head.next
    pre = None
    hash_list = list()

    while cur != None:
        if cur.data in hash_list:
            pre.next = cur.next
            cur = pre.next
        else:
            hash_list.append(cur.data)
            pre = cur
            cur = cur.next


if __name__ == "__main__":
    i = 1
    head = LNode(0)
    head.next = None
    tmp = None
    cur = head
    while i < 7:
        tmp = LNode(i)
        if i%2 == 0:
            tmp.data = i+1
        elif i%3 == 0:
            tmp.data = i-2
        else:
            tmp.data = i
        tmp.next = None
        cur.next = tmp
        cur = tmp
        i += 1
    print("删除重复节点前：")
    cur = head.next
    while cur != None:
        print(cur.data)
        cur = cur.next
    removeDup2(head)
    print("删除节点后：")
    cur = head.next
    while cur != None:
        print(cur.data)
        cur = cur.next

代码执行结果：

删除重复节点前：
1
3
1
5
5
7
删除节点后：
1
3
5
7