例3 一个字符串只包含*和数字，请把它的*号都放开头。


方法1：
快排partition——数字相对顺序会变化
循环不变式：[0..i – 1]都是*,[i..j– 1]是数字，[j..n– 1]未探测

程序如下：

#include <iostream>
#include <string.h>
#include <vector>
using namespace std;


class Solution {
public:

    char* Repace(char A[]) {
      int lenA = strlen(A);
      int i = 0;
      for(int j = 0;j < lenA; ++j)
      {
		  if(A[j] == '*')  
		  {
			  char temp = A[i];
			  A[i] = A[j];
			  A[j] = temp;
			  i++;
			}
	}
	return A;
}
};


int main()
{
	Solution* p = new Solution;
	char B[100] = "*1*2*3";
	//p->changeSpace(B);
	cout <<p->Repace(B)<< endl;
	delete p;
	return 0;
}

方法2 ：倒序，可保持数字位置相对不变
代码如下：

#include <iostream>
#include <string.h>
#include <vector>
using namespace std;


class Solution {
public:
char* daozhe(char A[])
{
	int lenA = strlen(A);
	int j = lenA - 1;
	for(int i = lenA - 1; i >= 0;--i)
	{
		if(A[i] != '*')
			A[j--] = A[i];
	}
	cout << j << endl;
	for(int i = 0;i <= j;++i)
		A[i] = '*';
	return A;
}
};


int main()
{
	Solution* p = new Solution;
	char B[100] = "*1*2*3";
	//p->changeSpace(B);
	cout <<p->daozhe(B)<< endl;
	delete p;
	return 0;
}