题解 | #逆波兰表达式求值#

/**用一个栈,数字就入，运算符就取两个计算
 * 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 <stdio.h>
#include <string.h>
int evalRPN(char** tokens, int tokensLen ) {
    int stack[10000];
    int point = 0;
    for (int i = 0; i < tokensLen; i++) {
        if (!strcmp(tokens[i], "+")) {
        
            stack[point - 2] = stack[point - 2] + stack[point - 1];
            point = point - 1;
        } else if (!strcmp(tokens[i], "-")) {
            stack[point - 2] = stack[point - 2] - stack[point - 1];
            point = point - 1;
        } else if (!strcmp(tokens[i], "*")) {
            stack[point - 2] = stack[point - 2] * stack[point - 1];
            point = point - 1;
        } else if (!strcmp(tokens[i], "/")) {
            stack[point - 2] = stack[point - 2] / stack[point - 1];
            point = point - 1;
        } else {
            stack[point] = atoi(tokens[i]);
            point++;
        }

    }

    return stack[0];
}