【算法day8】整数反转

整数反转

https://leetcode.cn/problems/reverse-integer/description/

class Solution {
public:
    int reverse(int x) {
        int MAX_LENGTH = 11; // 32位整数的最大数字的位数
        int* num = (int*)calloc(sizeof(int), MAX_LENGTH); //用于保存进位每一位的数字
        int current = x;
        int pos = 0;
        while (current != 0 && pos < MAX_LENGTH) {
            num[pos] = current % 10;
            current = current / 10;
            pos++;
        }
        if (pos == MAX_LENGTH) {
            return 0;
        }
        int new_pos = 1;
        int new_num = num[0];
        int highest_carry = 10;
        while (new_pos < pos) {
            long after = new_num;
            after = after * 10 + num[new_pos];
            // 若进位发生了变化，那么返回0，说明出现了溢出，
            if (after % highest_carry != num[new_pos]) {
                return 0;
            } else {
                new_num = after;
            }
            if (after == 0) {
                highest_carry *= 10;
            } else {
                highest_carry = after * 10;
            }
            new_pos++;
        }
        return new_num;
    }
};