Other algorithms -002- replace spaces

Title Description

Please implement a function, replace a string into each space “%20”. For example, when the string is We Are Happy.the string through replaced afterWe%20Are%20Happy

analysis

• A thought: 从头到尾traversal strings do replacement time complexity is O(n2)inefficient.
  

• Two ideas: string made from the tail to the head traversal Alternatively, time complexity is O (n), and high efficiency.
  

• Python use strthe built-in `method, easy to solve. Time and space complexity is not optimal.

Code

// C++
class Solution {
public:
    // 指向字符数组的字符指针str，字符数组长度length
	void replaceSpace(char *str,int length) {

        // 边界检查1：判断字符数组是否为空
	    if(str==NULL)
            return ;
        // 遍历字符串，统计空格个数、替换前字符个数、替换后字符个数
        int CountOfBlanks=0; // 空格个数
        int Originallength=0;// 替换前字符个数
        int len=0;           // 替换后字符个数

        for(int i=0;str[i]!='\0';++i)
        {
            Originallength++;
            if(str[i]==' ')
                ++CountOfBlanks;
        }

        len =Originallength+2*CountOfBlanks;

        // 边界检查2：判断字符数组是否越界
        if(len+1>length)
             return ;

        // 替换空格
        char*pStr1=str+Originallength;// 字符指针指向原始字符串的末尾
        char*pStr2=str+len;           // 字符指针指向替换后字符串的末尾

        while(pStr1 != pStr2)         // 替换结束的条件
        {
            if(*pStr1==' ')
            {
                *pStr2--='0';
                *pStr2--='2';
                *pStr2--='%';
            }
            else
            {
                *pStr2--=*pStr1;
            }
            --pStr1;
        }
	}
};

# python
# -*- coding:utf-8 -*-
class Solution:
    # s 源字符串
    def replaceSpace(self, s):
        # write code here
        return '%20'.join(s.split(' '))
        #return s.replace(' ', '%20')