【LeetCode】7. Reverse Integer 解題報告

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7. Reverse Integer / Easy

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Example 4:

Input: x = 0
Output: 0

Constraints:

  • -2^31 <= x <= 2^31 - 1

Solution 1: Brute Force

思路

這題算是很簡單，只是由於環境不允許存 64bit，所以轉換過程中可能會 overflow，為了避免這個狀況，把原來的數字不斷除十，將每次的個位數存入 vector，之後再迴圈一個一個加乘回來。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    int reverse(int x) {
        bool sign = x < 0 ? true : false;
        vector<int> reversed_int;
        x = abs(x);
        while(x > 0) {
            int tmp = x%10;
            reversed_int.emplace_back(tmp);
            x /= 10;
        }
        
        long long int sum = 0;
        for(int i = 0; i < reversed_int.size(); ++i) {
            sum += reversed_int[i] * pow(10, reversed_int.size() - i - 1);
        }

        if(sum > INT_MAX) return 0;
        if(sum < INT_MIN) return 0;
        if(sign) return sum*-1;
        return sum;
    }
};