C语言 定义一个函数实现两个集合的相加运算。(用链表的方式实现)
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发布时间：2023-07-19 15:44
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时间：2024-11-23 13:50
#include #include #include #define N 7 typedef enum { add, nul, sub, div1, yu, l, r }OP; int a[N][N] = { { 0, 0, -1, -1, -1, 1, 2 }, { 0, 0, -1, -1, -1, 1, 2 }, { 1, 1, 0, 0, 0, 1, 2 }, { 1, 1, 0, 0, 0, 1, 2 }, { 1, 1, 0, 0, 0, 1, 2 }, }; int top; OP beg; int b[1024]; OP op[1024]; void init_nu( ){ top = 0; } void push_nu( int term ){ b[top++] = term; } int pop_nu( ){ return b[--top]; } int is_empty_nu( ){ return top == 0; } void destory_nu( ) { top = 0; } void init_sign( ){ beg = 0; } void push_sign( OP sign ){ op[beg++] = sign; } void destory_sign( ){ beg = 0; } OP pop_sign( ){ return op[--beg];} OP get_sign( ){ return op[beg - 1]; } int is_empty_sign( ){return beg == 0; } int eval() { int i, j; i = pop_nu(); j = pop_nu(); switch( pop_sign() ) { case '+': push_nu( j + i ); break; case '-': push_nu( j - i ); break; case '*': push_nu( j * i ); break; case '/': push_nu( j / i ); break; case '%': push_nu( j & i ); break; defult: break; } } int change( char *s ) { int i; int n = strlen( s ); for( i = 0; i n; i++ ) { if( s[i] >= '0' && s[i] = '0' && s[i] <= '9' ) push_nu( 10 * pop_nu() + s[i++] - '0' ); switch( s[i] ) { case '+': while( a[add][get_sign()] <= 0 ) eval(); push_sign( add ); break; case '-': while( a[nul][get_sign()] <= 0 ) eval(); push_sign( nul ); break; case '*': while( a[sub][get_sign()] <= 0 ) eval(); push_sign( sub ); break; case '/': while( a[div1][get_sign()] <= 0 ) eval(); push_sign( div1 ); break; case '%': while( a[yu][get_sign()] <= 0 ) eval(); push_sign( yu ); break; case '(': push_sign( l ); break; case ')': while( (get_sign()) != l ) eval(); pop_sign(); break; defult: break; } } return pop_nu(); } int main( void ) { char *s = "((5-3)*2+4/2&2+1)"; init_nu(); init_sign(); printf( "%d\n", change( s )); destory_nu(); destory_sign(); return 0; } 以前编的，希望对你有帮助。别忘了给我分。
