算法百题004：数据压缩——游程编码

题目：给定一个包含ASCII码的字符串，利用下列规则进行编码

1）将子串中连续k个相同字符c编为kc

2）若子串长度为1，则1忽略。

 

实现一：整体编码完再输出

char * encoding_v1( char * str_in ){char * str_out = new char[100];memset( str_out, '/0', sizeof(char) * 100 );int freq = 1;char num[10] = { '/0' };for( int i = 0; str_in[i] != '/0'; ++i ){if( str_in[i] == str_in[i+1] ){++freq;}else{if( freq > 1 ){itoa( freq, num, 10 );strcat( str_out, num );}char s[] = { str_in[i], '/0'};strcat( str_out, s );freq = 1;}}return str_out;}

 

实现二：编码的同时逐步输出结果

void encoding_v2( char * str ){int freq = 1;for( int i = 0; str[i] != '/0'; ++i ){if( str[i] == str[i+1] ){++freq;}else{if( freq > 1 ){cout << freq;}cout << str[i];freq = 1;}}cout << endl;}

 

实现三：利用std::string类实现

void encode_v3( const string & s ){int freq = 1;for( int j = 0; j < s.size(); j++ ){if( s[j] == s[j+1] )++freq;else{cout << freq << s[j];freq = 1;}}cout << endl;}

 

实现四：不是相邻两元素比较，而是与新串首字符比较。

void encode_v4( const string & s ){string t;t = s[0];int freq = 0;for( int i = 0; i < s.size(); ++i ){if( s[i] == t[0] ){++freq;}else{printf( "%d%c", freq, t[0] );t[0] = s[i];freq = 1;}if( i == s.size() - 1 )printf("%d%c", freq, t[0] );}printf("/n");}