lime1991的专栏

小硕即将面临找工作的压力，没事编编Leetcode。分享一些经验大家。刚开始，博主低估了LeetCode的要求，采用了一个最直观偷懒的方法。两层循环遍历，时间复杂度O(N^2)。LeetCode返回结果Time Limited Exceeded。贴出挫代码引以为戒。class Solution {public: vector twoSum(vector &numbers, i

/从链表的头结点开始，若phead1>phead2,那么合并后链表的头结点是phead1.反之，则是phead2//剩余链表仍然有序，合并步骤同上一步一样，可采用递归的方法解决//边界条件 phead1为NULL，phead2为NULL#include<iostream>using namespace std;struct ListNode{ int value; ListN

//两棵二叉树判断B是不是A的子结构//1.在A中遍历，找到节点与B根节点的值相等的节点//2.对该节点和B树遍历，判断结构是否一样//递归#include<iostream>using namespace std;struct Node{ int value; Node* left; Node* right;};bool Traver(Node* pa, N

//如果是二叉搜索树的话，//若当前节点t大于节点u，v，说明u，v的LCA在t的左节点中，故从t的左节点中继续查找//若当前节点t小于节点u，v，说明u，v的LCA在t的右节点中，从t的右节点继续查找//当找到一个节点满足u<t<v时，该节点就是u，v的LCA//当t的一个节点的值等于u或v时，说明其中一个节点是另一个节点的父节点，返回t的父节点struct TreeNode{

#include<iostream>using namespace std;//1.先判断是否有环，快慢指针，慢指针一次向后前进一格，快指针一次向后前进两格（注意判断是否为NULL）//当快指针为NULL时，两指针仍为相遇则不存在环。//2.判断环中包含元素的个数。//以两指针相遇处为初始点，慢指针依次往后挪，并计数，当再次回到相遇点时，可求得包含元素个数n。//3.找出环的入口。指针p1

题目描述：Given two sorted integer arrays nums1 andnums2, merge nums2 into nums1 as one sorted array.Note:You may assume that nums1 has enough space(size that is greater or equal to m + n) to hold ad

题目描述：You are given two linked lists representingtwo non-negative numbers. The digits are stored in reverse order and each oftheir nodes contain a single digit. Add the two numbers and return it as a

题目描述：Given an array of integers, every elementappears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without usi

问题描述：Implement atoi to convert a string to an integer.Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible i

题目描述：There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).解法：常规解法是将nums1和n

问题描述：Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is

转载自http://blog.youkuaiyun.com/cumirror/article/details/5596916/* * ************************************************************************************ *       hash_map映照容器的基础说明： *************

/*深度优先搜索用栈（stack）来实现，整个过程可以想象成一个倒立的树形：1、把根节点压入栈中。2、每次从栈中弹出一个元素，搜索所有在它下一级的元素，把这些元素压入栈中。并把这个元素记为它下一级元素的前驱。3、找到所要找的元素时结束程序。4、如果遍历整个树还没有找到，结束程序。广度优先搜索使用队列（queue）来实现，整个过程也可以看做一个倒立的树形：1、把根节点放到队列的末尾。2