LeetCode 0246. Strobogrammatic Number Solution in Java, C++, Python & Go | Explanation + Code

Description

Given a string num which represents an integer, return true if num is a strobogrammatic number.

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

 

Example 1:

Input: num = "69"
Output: true

Example 2:

Input: num = "88"
Output: true

Example 3:

Input: num = "962"
Output: false

 

Constraints:

• 1 <= num.length <= 50
• num consists of only digits.
• num does not contain any leading zeros except for zero itself.

Solutions

Solution 1: Two Pointers Simulation

We define an array d, where d[i] represents the number after rotating the digit i by 180°. If d[i] is -1, it means that the digit i cannot be rotated 180° to get a valid digit.

We define two pointers i and j, pointing to the left and right ends of the string, respectively. Then we continuously move the pointers towards the center, checking whether d[num[i]] and num[j] are equal. If they are not equal, it means that the string is not a strobogrammatic number, and we can directly return false. If i > j, it means that we have traversed the entire string, and we return true.

The time complexity is O(n), where n is the length of the string. The space complexity is O(1).

PythonJavaC++Go

class Solution: def isStrobogrammatic(self, num: str) -> bool: d = [0, 1, -1, -1, -1, -1, 9, -1, 8, 6] i, j = 0, len(num) - 1 while i <= j: a, b = int(num[i]), int(num[j]) if d[a] != b: return False i, j = i + 1, j - 1 return True(code-box)

class Solution { public boolean isStrobogrammatic(String num) { int[] d = new int[] {0, 1, -1, -1, -1, -1, 9, -1, 8, 6}; for (int i = 0, j = num.length() - 1; i <= j; ++i, --j) { int a = num.charAt(i) - '0', b = num.charAt(j) - '0'; if (d[a] != b) { return false; } } return true; } }(code-box)

class Solution { public: bool isStrobogrammatic(string num) { vector<int> d = {0, 1, -1, -1, -1, -1, 9, -1, 8, 6}; for (int i = 0, j = num.size() - 1; i <= j; ++i, --j) { int a = num[i] - '0', b = num[j] - '0'; if (d[a] != b) { return false; } } return true; } };(code-box)

func isStrobogrammatic(num string) bool { d := []int{0, 1, -1, -1, -1, -1, 9, -1, 8, 6} for i, j := 0, len(num)-1; i <= j; i, j = i+1, j-1 { a, b := int(num[i]-'0'), int(num[j]-'0') if d[a] != b { return false } } return true }(code-box)