In this video, we solve LeetCode Problem 7: Reverse Integer using C++.
We are given a signed 32-bit integer x, and we need to reverse its digits. If the reversed integer goes outside the 32-bit signed integer range, we return 0.
In this solution, we cover:
- How to extract the last digit using modulo %
- How to build the reversed number step by step
- How to handle negative numbers
- How to check integer overflow without using 64-bit integers
- Time and space complexity explanation
Approach:
We repeatedly take the last digit of the number using:
last = x % 10;
Then we add it to the reversed number:
rev = rev * 10 + last;
Before multiplying by 10, we check whether rev may overflow the 32-bit integer range.
Complexity:
Time Complexity: O(log10(n))
Space Complexity: O(1)
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Tags:
LeetCode 7 Reverse Integer C++ LeetCode C++ Solution Integer Overflow DSA Coding Interview C++ LeetCode Reverse Number
