中缀表达式转前缀或后缀表达式
前缀(后缀)表达式可以在不使用"()"时保证表达式不产生歧义,更适合计算机计算。
此处以后缀为例:
*a+b = ab+ a-b*c = abc -
方法:
1.创建栈
2.从左向右顺序获取中缀表达式
a.数字直接输出
b.运算符
情况一:遇到左括号直接入栈,遇到右括号将栈中左括号之后入栈的运算符全部弹栈输出,同时左括号出栈但是不输出。
情况二:遇到乘号和除号直接入栈,直到遇到优先级比它更低的运算符,依次弹栈。
情况三:遇到加号和减号,如果此时栈空,则直接入栈,否则,将栈中优先级高的运算符依次弹栈(注意:加号和减号属于同一个优先级,所以也依次弹栈)直到栈空或则遇到左括号为止,停止弹栈。(因为左括号要匹配右括号时才弹出)。
情况四:获取完后,将栈中剩余的运算符号依次弹栈输出
例:比如将:2*(9+6/3-5)+4转化为后缀表达式 2 9 6 3 / +5 - * 4 +
中缀转后缀代码如下
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| #include <stdio.h> #include <stdlib.h> #include <ctype.h> #include <string.h>
#define MAX 100
char stack[MAX]; int top = -1;
int precedence(char op) { if (op == '+' || op == '-') { return 1; } else if (op == '*' || op == '/') { return 2; } return 0; }
int isOperator(char ch) { return ch == '+' || ch == '-' || ch == '*' || ch == '/'; }
void push(char op) { if (top < MAX - 1) { stack[++top] = op; } else { printf("Stack overflow\n"); } }
char pop() { if (top != -1) { return stack[top--]; } else { printf("Stack underflow\n"); return '\0'; } }
char peek() { if (top != -1) { return stack[top]; } return '\0'; }
void infixToPostfix(char *infix) { char postfix[MAX] = {0}; int i, j = 0; for (i = 0; i < strlen(infix); i++) { char ch = infix[i];
if (isdigit(ch) || isalpha(ch)) { postfix[j++] = ch; } else if (ch == '(') { push(ch); } else if (ch == ')') { while (top != -1 && peek() != '(') { postfix[j++] = pop(); } pop(); } else if (isOperator(ch)) { while (top != -1 && precedence(peek()) >= precedence(ch)) { postfix[j++] = pop(); } push(ch); } }
while (top != -1) { postfix[j++] = pop(); }
printf("Postfix: %s\n", postfix); }
int main() { char infix[MAX]; printf("Enter infix expression: "); scanf("%s", infix); infixToPostfix(infix); return 0; }
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中缀转前缀代码如下
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| #include <stdio.h> #include <stdlib.h> #include <ctype.h> #include <string.h>
#define MAX 100
char stack[MAX]; int top = -1;
int precedence(char op) { if (op == '+' || op == '-') { return 1; } else if (op == '*' || op == '/') { return 2; } return 0; }
int isOperator(char ch) { return ch == '+' || ch == '-' || ch == '*' || ch == '/'; }
void push(char op) { if (top < MAX - 1) { stack[++top] = op; } else { printf("Stack overflow\n"); } }
char pop() { if (top != -1) { return stack[top--]; } else { printf("Stack underflow\n"); return '\0'; } }
char peek() { if (top != -1) { return stack[top]; } return '\0'; }
void reverse(char *exp) { int len = strlen(exp); for (int i = 0; i < len / 2; i++) { char temp = exp[i]; exp[i] = exp[len - i - 1]; exp[len - i - 1] = temp; } }
void infixToPrefix(char *infix) { char prefix[MAX] = {0}; int len = strlen(infix); reverse(infix); for (int i = 0; i < len; i++) { if (infix[i] == '(') { infix[i] = ')'; } else if (infix[i] == ')') { infix[i] = '('; } } int j = 0; for (int i = 0; i < len; i++) { char ch = infix[i];
if (isdigit(ch) || isalpha(ch)) { prefix[j++] = ch; } else if (ch == '(') { push(ch); } else if (ch == ')') { while (top != -1 && peek() != '(') { prefix[j++] = pop(); } pop(); } else if (isOperator(ch)) { while (top != -1 && precedence(peek()) > precedence(ch)) { prefix[j++] = pop(); } push(ch); } }
while (top != -1) { prefix[j++] = pop(); }
reverse(prefix);
printf("Prefix: %s\n", prefix); }
int main() { char infix[MAX]; printf("Enter infix expression: "); scanf("%s", infix); infixToPrefix(infix); return 0; }
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