题解 | #逆波兰表达式求值#

c语言版本

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param tokens string字符串一维数组
 * @param tokensLen int tokens数组长度
 * @return int整型
 */
#include <stdlib.h>
#include <string.h>
int evalRPN(char** tokens, int tokensLen ) {
    int stack[tokensLen];
    int top = -1, i = 0;
    for (i = 0; i < tokensLen; i++) {
        int res = 0;
        if (strcmp(tokens[i], "+") == 0) {
            res = stack[top] + stack[top - 1];
            top--;
            stack[top] = res;
        } else if (strcmp(tokens[i], "-") == 0) {
            res = stack[top - 1] - stack[top];
            top--;
            stack[top] = res;
        } else if (strcmp(tokens[i], "*") == 0) {
            res = stack[top] * stack[top - 1];
            top--;
            stack[top] = res;
        } else if (strcmp(tokens[i], "/") == 0) {
            res = stack[top - 1] / stack[top];
            top--;
            stack[top] = res;
        } else {
            stack[++top] = atoi(tokens[i]);
        }
    }
    return stack[top];
}