將字符串中的空格替換，時間複雜度爲0（n）解法

題目：實現一個函數，把字符串中的每個空格替換成“%20”。例如輸入“We are happy.”,則輸出”We%20are%20happy.”

題目很簡單，直接上代碼

#include<iostream>
using namespace std;

void Replace(char string[],int length)
{
    if(string==NULL || length<0)
    {
        return;
    }
    int originalLength=0;
    int newLength     =0;
    int numberofBlank =0;
    int i=0;
    while(string[i]!='\0')
    {
        ++originalLength;
        if(string[i]==' ')
        {
           ++numberofBlank;
        }
        ++i;
    }
    newLength=originalLength+numberofBlank*2;
    int indexofOriginal=originalLength;
    int indexNew=newLength;
    if(newLength>length)
    {
        return;
    }
    while(indexofOriginal>0 && indexofNew>indexofOriginal)
    {
        if(string[indexofOriginal]==' ')
        {
            string[indexofNew--]='0';
            string[indexofNew--]='2';
            string[indexofNew--]='%';
        }
        else{
            string[indexofNew--]=string[indexofOriginal];
        }
        --indexofOriginal;
    }

}

int main()
{
    char string1[50]="Nothing is impossible!";
    char string2[50]="My name is CSDN.";
    char string3[50]="We are happy.";
    Replace(string1,50);
    Replace(string2,50);
    Replace(string3,50);
    cout<<string1<<endl;
    cout<<string2<<endl;
    cout<<string3<<endl;
    system("pause");
    return 0;
}