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2544. Alternating Digit Sum

Description

You are given a positive integer n. Each digit of n has a sign according to the following rules:

• The most significant digit is assigned a positive sign.
• Each other digit has an opposite sign to its adjacent digits.

Return the sum of all digits with their corresponding sign.

 

Example 1:

Input: n = 521
Output: 4
Explanation: (+5) + (-2) + (+1) = 4.

Example 2:

Input: n = 111
Output: 1
Explanation: (+1) + (-1) + (+1) = 1.

Example 3:

Input: n = 886996
Output: 0
Explanation: (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.

 

Constraints:

• 1 <= n <= 109

 

Solutions

Solution 1: Simulation

We can directly simulate the process as described in the problem.

We define an initial symbol $sign=1$. Starting from the most significant digit, we take out one digit $x$ each time, multiply it by $sign$, add the result to the answer, then negate $sign$, and continue to process the next digit until all digits are processed.

The time complexity is $O(\log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the given number.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int alternateDigitSum(int n) {
          int ans = 0, sign = 1;
          for (char c : String.valueOf(n).toCharArray()) {
              int x = c - '0';
              ans += sign * x;
              sign *= -1;
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int alternateDigitSum(int n) {
          int ans = 0, sign = 1;
          for (char c : to_string(n)) {
              int x = c - '0';
              ans += sign * x;
              sign *= -1;
          }
          return ans;
      }
  };
  

• class Solution:
      def alternateDigitSum(self, n: int) -> int:
          return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))
  
  

• func alternateDigitSum(n int) (ans int) {
  	sign := 1
  	for _, c := range strconv.Itoa(n) {
  		x := int(c - '0')
  		ans += sign * x
  		sign *= -1
  	}
  	return
  }
  

• function alternateDigitSum(n: number): number {
      let ans = 0;
      let sign = 1;
      while (n) {
          ans += (n % 10) * sign;
          sign = -sign;
          n = Math.floor(n / 10);
      }
      return ans * -sign;
  }
  
  

• impl Solution {
      pub fn alternate_digit_sum(mut n: i32) -> i32 {
          let mut ans = 0;
          let mut sign = 1;
          while n != 0 {
              ans += (n % 10) * sign;
              sign = -sign;
              n /= 10;
          }
          ans * -sign
      }
  }
  
  

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