<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[lanse20的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/lanse20</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; lanse20]]></copyright><item><title><![CDATA[【转载】梯度下降法（一）入门]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78629456</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78629456</guid><author>lanse20</author><pubDate>Sat, 25 Nov 2017 00:50:39 +0800</pubDate><description><![CDATA[原文请看这里：http://www.52ml.net/2038.html

梯度下降法
是一个

一阶
最优化算法
，通常也称为
最速下降法
。我之前也没有关注过这类算法。最近，听斯坦福大学的机器学习课程时，碰到了用梯度下降算法求解线性回归问题，于是看了看这类算法的思想。今天只写了一些入门级的知识。







我们知道，函数的曲线如下： 



编程实]]></description><category></category></item><item><title><![CDATA[c++ 用两个栈来实现一个队列，完成队列的Push和Pop操作。 队列中的元素为int类型。]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78504406</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78504406</guid><author>lanse20</author><pubDate>Sat, 11 Nov 2017 03:01:54 +0800</pubDate><description><![CDATA[//参考了很多网上的代码。。。不是原创
class Solution
{
public:
    void push(int node) {
        stack1.push(node);
    }

    int pop() {
        int result=0;
        int temp=0;
        if(stack2.empty())
        {]]></description><category></category></item><item><title><![CDATA[c++ 输入一个链表，从尾到头打印链表每个节点的值。]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78504397</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78504397</guid><author>lanse20</author><pubDate>Sat, 11 Nov 2017 02:45:56 +0800</pubDate><description><![CDATA[//这个题默认链表的头结点不为空，是有数值的。第一次做的时候，以为/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Sol]]></description><category></category></item><item><title><![CDATA[c++ 请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78504381</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78504381</guid><author>lanse20</author><pubDate>Sat, 11 Nov 2017 01:48:00 +0800</pubDate><description><![CDATA[//注意字符串最后一个字符为'\0'
class Solution {
public:
	void replaceSpace(char *str,int length) {
        int blank_len=0;
        int old_len=0,new_len=0;
        int i=0;
        while(str[i]!='\0')
        {]]></description><category></category></item><item><title><![CDATA[8. String to Integer (atoi)  LeetCode c++]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78336412</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78336412</guid><author>lanse20</author><pubDate>Wed, 25 Oct 2017 00:16:10 +0800</pubDate><description><![CDATA[//已经提交，若有不足之处，请留言，非常感谢！！！
//#include 
//using namespace std;

class Solution {
public:
    int myAtoi(string str) {
        int len_str=str.length();
        if(str=="")
            return INT_MAX;
 ]]></description><category></category></item><item><title><![CDATA[7. Reverse Integer LeetCode c++]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78335845</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78335845</guid><author>lanse20</author><pubDate>Tue, 24 Oct 2017 22:44:03 +0800</pubDate><description><![CDATA[//可以提交。有不足之处请留言，谢谢！
class Solution {
public:
    int reverse(int x) {
        if((x>(1<<31-1))||(x<-(1<<31))) //范围判断
            return 0;
        int out_x=0;
        int temp_x=abs(x);
        int ]]></description><category></category></item><item><title><![CDATA[6. ZigZag Conversion leetcode]]></title><link>https://blog.youkuaiyun.com/lanse20/article/details/78335268</link><guid>https://blog.youkuaiyun.com/lanse20/article/details/78335268</guid><author>lanse20</author><pubDate>Tue, 24 Oct 2017 21:51:12 +0800</pubDate><description><![CDATA[//这个可以提交。有问题或者建议可以留言，谢谢！
class Solution {
public:
    string convert(string s, int numRows) {
        string out_s="";
        int size_s=s.length();
        int step=0;//每个zigzag包含的字符个数
        int ]]></description><category></category></item></channel></rss>