pengyu2003

摘要： Given a linked list, remove thenthnode from the end of list and return its head.For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.Note:Givennwill always be valid.Try to do this in one pass.各种 阅读全文

摘要： Given a string containing just the characters'(',')','{','}','['and']', determine if the input string is valid.The brackets must close in the correct order,"()"and"()[]{}"are all valid but"(]"and"([)]"are not.用栈会 阅读全文

摘要： Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.if... else if不要偷懒直接写 if...if.../** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x)... 阅读全文

摘要： Given a linked list, swap every two adjacent nodes and return its head.For example,Given1->2->3->4, you should return the list as2->1->4->3.Your algorithm should use only constant space. You maynotmodify the values in the list, only nodes itself can be changed.代码不是最简的，处理head的部分明显可以 阅读全文

摘要： 今天刷了4道题。总结：1.==------>=不要笔误2.该if...elseif...的不要只写if...if...可能上个if运行完继续下个if。3.vector随机读很快，但随机删除很慢。定位的时候，如果用迭代器iterator比较麻烦，如果删除vector的第i个元素，不妨直接（vectorobj）.erase(obj.begin()+i,obj.begin()+i+1);其中删除的是前面的元素，最后一个元素不删除。欢欢乐乐看书去了~~ 阅读全文