Leetcode 7. Reverse Integer

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

• Two loops

class Solution {
public:
    int reverse(int x) {
        if(x == 0 || x == INT_MIN) {
            return 0;
        }
        int remainder = 0;
        int y = abs(x);
        queue<int> number;
        while(y) {
            remainder = y % 10;
            number.push(remainder);
            y /= 10;
        }
        int result = 0;
        while(!number.empty()) {
            if((INT_MAX - number.front()) / 10 < result) {
                return 0;
            }
            result = result * 10 + number.front();
            number.pop();
        }
        return x == abs(x)?result:-result;
    }
};

• One loop

  class Solution {
  public:
      int reverse(int x) {
          if(x == 0 || x == INT_MIN) {
              return 0;
          }
          int y = abs(x);
          int result = 0;
          int remainder = 0;
          while(y) {
              remainder = y % 10;
              if((INT_MAX - remainder) / 10 < result) {
                  return 0;
              }
              result = result * 10 + remainder;
              y /= 10;
          }
          return x == abs(x)?result:-result;
      }
  };

Reference

1. https://leetcode.com/problems/reverse-integer/description/