题目链接:https://leetcode-cn.com/problems/reverse-integer/
题意:整数反转
分析:有点麻烦,因为控制了边界,还得判断反转后是否超出了32位的范围。
class Solution { public: int reverse(int x) { int rev = 0; while (x != 0) { int pop = x % 10; x /= 10; if (rev > INT_MAX/10 || (rev == INT_MAX / 10 && pop > 7)) return 0; if (rev < INT_MIN/10 || (rev == INT_MIN / 10 && pop < -8)) return 0; rev = rev * 10 + pop; } return rev; } };