https://blog.csdn.net/m_nValuezh-cnhttps://blog.csdn.net/https://blog.csdn.net/m_nValue/article/details/52540253https://blog.csdn.net/m_nValue/article/details/52540253m_nValueWed, 14 Sep 2016 17:26:03 +0800 function ascii2native(){ var character=document.getElementById("src").value.split("\\u");]]>https://blog.csdn.net/m_nValue/article/details/47083715https://blog.csdn.net/m_nValue/article/details/47083715m_nValueMon, 27 Jul 2015 13:26:23 +0800https://blog.csdn.net/m_nValue/article/details/40683281https://blog.csdn.net/m_nValue/article/details/40683281m_nValueSat, 01 Nov 2014 19:48:54 +0800https://blog.csdn.net/m_nValue/article/details/39674603https://blog.csdn.net/m_nValue/article/details/39674603m_nValueMon, 29 Sep 2014 20:01:58 +0800https://blog.csdn.net/m_nValue/article/details/37596785https://blog.csdn.net/m_nValue/article/details/37596785m_nValueWed, 09 Jul 2014 13:26:53 +0800https://blog.csdn.net/m_nValue/article/details/36684411https://blog.csdn.net/m_nValue/article/details/36684411m_nValueThu, 03 Jul 2014 18:37:44 +0800https://blog.csdn.net/m_nValue/article/details/8140979https://blog.csdn.net/m_nValue/article/details/8140979m_nValueFri, 02 Nov 2012 15:56:45 +0800= 0;--t) { //y]]>https://blog.csdn.net/m_nValue/article/details/6359365https://blog.csdn.net/m_nValue/article/details/6359365m_nValueSun, 24 Apr 2011 20:55:00 +0800https://blog.csdn.net/m_nValue/article/details/6328934https://blog.csdn.net/m_nValue/article/details/6328934m_nValueSun, 17 Apr 2011 11:05:00 +0800https://blog.csdn.net/m_nValue/article/details/6313478https://blog.csdn.net/m_nValue/article/details/6313478m_nValueSun, 10 Apr 2011 14:30:00 +0800几种排序算法的时间、空间复杂度：


像你这样出色的男人，无论在什么地方，都像黑夜中的萤火虫一样，那样的鲜明，那样的出众。你那忧郁的眼神，稀疏的胡茬子，神乎其神的刀法，太阳啊~]]>https://blog.csdn.net/m_nValue/article/details/6310451https://blog.csdn.net/m_nValue/article/details/6310451m_nValueFri, 08 Apr 2011 19:45:00 +0800感谢XX公司的笔试啊，让我总算下定决心彻底把这个似乎很简单的问题想一想了，发觉之前自己的理解真是差强人意！
 
下面的链表翻转是通过重新创建一个新链表来实现的，空间复杂度过于不理想...
#include using namespace std; struct Node { int data; Node *next; }*head; void AddNode(int value); void Swap(Node *&head]]>https://blog.csdn.net/m_nValue/article/details/6306733https://blog.csdn.net/m_nValue/article/details/6306733m_nValueThu, 07 Apr 2011 12:07:00 +0800https://blog.csdn.net/m_nValue/article/details/6305534https://blog.csdn.net/m_nValue/article/details/6305534m_nValueWed, 06 Apr 2011 20:53:00 +0800程序没有实现与用户的输入交互：
#include using namespace std; void Sort(double [][2],int size); int main() { const int N=6; double max=0; double x1,y1,x2,y2; double pt[N][2]={{1.2,1.8},{2.1,8.7},{0.31,3.1},{1.4,4.9},{12.5,5.2},{6.8,6.1}};]]>https://blog.csdn.net/m_nValue/article/details/6283045https://blog.csdn.net/m_nValue/article/details/6283045m_nValueMon, 28 Mar 2011 11:18:00 +0800https://blog.csdn.net/m_nValue/article/details/6279908https://blog.csdn.net/m_nValue/article/details/6279908m_nValueSat, 26 Mar 2011 14:36:00 +0800//{2,1,2,3,5}不重复的元素个数是3
 
#include
#define N 15
using namespace std;

void Sort(int *,int size);

int main()
{
    bool flag=false;//以防出现连续重复的数值
    int k[N] = {1,17,1,3,]]>https://blog.csdn.net/m_nValue/article/details/6278708https://blog.csdn.net/m_nValue/article/details/6278708m_nValueFri, 25 Mar 2011 20:32:00 +0800>str;     strBuf=]]>https://blog.csdn.net/m_nValue/article/details/6262775https://blog.csdn.net/m_nValue/article/details/6262775m_nValueSun, 20 Mar 2011 14:49:00 +0800题目：
输入一个整形数组，数组中连续的一个或多个整数组成一个子数组，每个子数组都有一个和。
求所有子数组的和的最大值并输出该子数组成员。要求时间复杂度为O(n)。
#include
using namespace std;

int Sum(int *a,int size,int *&result,int &count);

int main()
{
]]>https://blog.csdn.net/m_nValue/article/details/6262741https://blog.csdn.net/m_nValue/article/details/6262741m_nValueSun, 20 Mar 2011 14:28:00 +0800