c语言-c语言编程基础之leetcode题解第7题整数反转.zip

#include <stdio.h> #include <stdlib.h> #include <limits.h> static int reverse(int x) { int n = 0; while (x != 0) { int r = x % 10; /* Checking the over/underflow. */ if (n > INT_MAX / 10 || (n == INT_MAX / 10 && r > 7)) { return 0; } if (n < INT_MIN / 10 || (n == INT_MIN / 10 && r < -8)) { return 0; } n = n * 10 + r; x /= 10; } return n; } #define TEST(n, e) printf("%12d => %-12d %s!\n", n, reverse(n), e == reverse(n)?"passed":"failed") int main(int argc, char**argv) { //basic cases TEST( 123, 321); TEST( -123, -321); TEST( -100, -1); TEST( 1002, 2001); //big integer TEST( 1463847412, 2147483641); TEST(-2147447412, -2147447412); TEST( 2147447412, 2147447412); //overflow TEST( 1000000003, 0); TEST( 2147483647, 0); TEST(-2147483648, 0); //customized cases if (argc != 2) { fprintf(stderr, "Usage: ./test i\n"); return 0; } printf("\n"); int n = atoi(argv[1]); printf("%12d => %-12d %s!\n", n, reverse(n), reverse(reverse(n))==n ? "passed":"failed"); return 0; }