C++ solution || Easy || O(n) || (95.17%)
```\nstring entityParser(string text) {\n        string s;\n        int i;\n        for(i=0;i
```\nstring entityParser(string text) {\n        string s;\n        int i;\n        for(i=0;i<text.length();i++){\n            if(text[i]==\'&\'){\n                if(text[i+1]==\'q\' && text[i+2]==\'u\' && text[i+3]==\'o\' && text[i+4]==\'t\' && text[i+5]==\';\'){\n                    s+="\\"";\n                    i+=5;\n                }else if(text[i+1]==\'a\' && text[i+2]==\'p\' && text[i+3]==\'o\' && text[i+4]==\'s\' && text[i+5]==\';\'){\n                    s+="\'";\n                    i+=5;\n                }else if(text[i+1]==\'a\' && text[i+2]==\'m\' && text[i+3]==\'p\' && text[i+4]==\';\'){\n                    s+="&";\n                    i+=4;\n                }else if(text[i+1]==\'f\' && text[i+2]==\'r\' && text[i+3]==\'a\' && text[i+4]==\'s\' && text[i+5]==\'l\' && text[i+6]==\';\'){\n                    s+="/";\n                    i+=6;\n                }else if(text[i+1]==\'g\' && text[i+2]==\'t\' && text[i+3]==\';\'){\n                    s+=">";\n                    i+=3;\n                }else if(text[i+1]==\'l\' && text[i+2]==\'t\' && text[i+3]==\';\'){\n                    s+="<";\n                    i+=3;\n                }else{\n                    s+=text[i];\n               }\n            }else{\n                s+=text[i];\n            }\n            }\n        return s;
['C']