sumIv1210的博客

其实这个题目有一个关键就是sorted array  如果不是sorted的话 可以用array index bit的类似方法解决这个问题 Remove duplicates from sorted array 1 1. Do not allocate extra space for another array, you must do this in place with constant

method 1 dummy node /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { pub

method 2. O(n2)的方法 class Solution { public: vector twoSum(vector& nums, int target) { vector result; for (int i = 0; i < nums.size(); ++i) { for (int j = i+1;

1. Palindrome linked list 这个解法很不错呀 利用了reverse linked list的方法 + 两个指针 class Solution { public: ListNode* reverse(ListNode* node) { ListNode* prev = NULL; while (node != NULL)

又是一个简单的但是都没一次通过的题目， 第一遍刷题就这么差么 自己丑陋的代码 so ugly class Solution { public: int reverse(int x) { if ((x >= 2147483647) || (x <= -2147483648) || x == 0) { return 0;

这个貌似复杂了点 丑陋的代码 class Solution { public: int removeDuplicates(vector& nums) { if (nums.size() < 3) { return nums.size(); } int c = 0; int i = 1

第一次出现了两次错误 class Solution { public: ListNode* oddEvenList(ListNode* head) { bool flag = false; if (head == NULL || head->next == NULL) { return head; }

1.dummy Node 的使用 感觉不是简单的就是 new一个dummy node 然后 dummy->next = head； 就好了， 这样是虚的 最关键的是 dummy->next = head; 之后 head 要再从dummy 开始 head = dummy； 这样的dummy才是有效的 并且连续的 2. reverse linked list 2 其实和1 很像 核心部

This question is just something like if you know the trick then you can do it, otherwise just no way to solve it. 两个指针 Note: 1. The faster one move 2 steps and slower one move one step each time 2

Remove duplicate in linked list 1 Note 1. when you want access n->next you need to make sure that n is not NULL, but when you want to access n->next->val, you need to make sure that n->next is not

这次又没有能一次通过....so sad..... solution 1: test case 最基本的[1，2，3，4，5，6] 没有能通过 然后想到了加一个flag  class Solution { public: ListNode* deleteDuplicates(ListNode* head) { if (head == NULL || head-

方法1 一次AC了 class Solution { public: void reorderList(ListNode* head) { if (head == NULL || head->next == NULL || head->next->next == NULL) { return; } L

dummy node的妙用 1.切记 用了dummy node以后 要head 从dummy开始，把他们连接起来 ListNode* dummy = new ListNode(0); dummy->next = head; head = dummy;2. 用temp然后delete temp 完成删除 class Solution { public: ListNode* remov