背单词太快就像龙卷风的博客

Sloution 1: # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseBetween(self, head: ListNode, left: int, right: int) -.

In a linknode list，we define a = list()，then we add some nodes，if we write a[0].next = a[1]， that's fine ,it works, but when we turn to a[1].next = a[0]，it time exceed，which confused me a lot. But finally I solved it. because,if we say a[1].next = a[0]...

Solution 1： It is a sorted linked list, we can delete all duplicates one by one,but at first I made a mistake in case [0,0,0,0,0] , because I set pre number 0,therefore I turn it to be -99,that is okay. class Solution { public: ListNode* deleteDu.

Tips:There is something noteworthy in C++ code，function getlen(LinkList * head)，this formal parameteris fake，if we call thisfuntion，ouractual parameter head will not change。That is different from array，for instance int fun(int a[])，it will change our a...