@@ -502,6 +502,67 @@ impl Solution {
502502}
503503```
504504
505+ ### C:
506+ 507+ ``` c
508+ int str_to_int (char * str) {
509+ // string转integer
510+ int num = 0, tens = 1;
511+ for (int i = strlen(str) - 1; i >= 0; i--) {
512+ if (str[ i] == '-') {
513+ num * = -1;
514+ break;
515+ }
516+ num += (str[ i] - '0') * tens;
517+ tens * = 10;
518+ }
519+ return num;
520+ }
521+ 522+ int evalRPN(char** tokens, int tokensSize) {
523+ 524+ int *stack = (int *)malloc(tokensSize * sizeof(int));
525+ assert(stack);
526+ int stackTop = 0;
527+ 528+ for (int i = 0; i < tokensSize; i++) {
529+ char symbol = (tokens[i])[0];
530+ if (symbol < '0' && (tokens[i])[1] == '0円') {
531+ 532+ // pop两个数字
533+ int num1 = stack[--stackTop];
534+ int num2 = stack[--stackTop];
535+ 536+ // 计算结果
537+ int result;
538+ if (symbol == '+') {
539+ result = num1 + num2;
540+ } else if (symbol == '-') {
541+ result = num2 - num1;
542+ } else if (symbol == '/') {
543+ result = num2 / num1;
544+ } else {
545+ result = num1 * num2;
546+ }
547+ 548+ // push回stack
549+ stack[stackTop++] = result;
550+ 551+ } else {
552+ 553+ // push数字进stack
554+ int num = str_to_int(tokens[i]);
555+ stack[stackTop++] = num;
556+
557+ }
558+ }
559+ 560+ int result = stack[0];
561+ free(stack);
562+ return result;
563+ }
564+ ```
565+
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