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本帖最后由 Danmoits 于 2021-1-1 15:00 编辑
测试输入:
2 1 + 3 *
预期输出:
9
解释:
该算式转化为常见的中缀算术表达式为:((2 + 1) * 3) = 9
- #include <stdio.h>
- #include <stdlib.h>
- #include <string.h>
- #include <ctype.h>
- typedef struct stack_t stack_t, *pstack_t, snode_t, *psnode_t;
- struct stack_t { psnode_t next; int v; };
- stack_t STACK(void);
- int isempty (pstack_t stack);
- int top (pstack_t stack);
- int pop (pstack_t stack);
- void push (pstack_t stack, int v);
- int main(void)
- {
- stack_t opnds = STACK();
- int operand, a, b;
- char input[32];
- while (scanf("%s", input) != EOF) {
- if (sscanf(input, "%d", &operand)) {
- push(&opnds, operand);
- } else {
- b = pop(&opnds);
- a = pop(&opnds);
- switch (input[0]) {
- case '+': push(&opnds, a + b); break;
- case '-': push(&opnds, a - b); break;
- case '*': push(&opnds, a * b); break;
- case '/': push(&opnds, a / b); break;
- }
- }
- }
- printf("%d", pop(&opnds));
- return 0;
- }
- stack_t STACK(void) { return (stack_t){ NULL }; }
- int isempty (pstack_t stack) { return stack->next == NULL; }
- int top (pstack_t stack) { return isempty(stack) ? -1 : stack->next->v; }
- int pop (pstack_t stack) {
- if (isempty(stack)) return -1;
- psnode_t p = stack->next;
- int v = p->v;
- stack->next = p->next;
- free(p);
- return v;
- }
- void push (pstack_t stack, int v) {
- psnode_t p = (psnode_t)malloc(sizeof(snode_t));
- p->v = v;
- p->next = stack->next;
- stack->next = p;
- }
复制代码
请简单说下思路,并加尽可能详细的注释,谢谢 |
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