Question

2710. Remove Trailing Zeros From a String

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Description

Given a positive integer num represented as a string, return _the integer _num_ without trailing zeros as a string_.

 

Example 1:

Input: num = "51230100"
Output: "512301"
Explanation: Integer "51230100" has 2 trailing zeros, we remove them and return integer "512301".

Example 2:

Input: num = "123"
Output: "123"
Explanation: Integer "123" has no trailing zeros, we return integer "123".

 

Constraints:

• 1 <= num.length <= 1000
• num consists of only digits.
• num doesn't have any leading zeros.

Solutions

Solution 1

class Solution:
  def removeTrailingZeros(self, num: str) -> str:
    return num.rstrip("0")

class Solution {
  public String removeTrailingZeros(String num) {
    int i = num.length() - 1;
    while (num.charAt(i) == '0') {
      --i;
    }
    return num.substring(0, i + 1);
  }
}

class Solution {
public:
  string removeTrailingZeros(string num) {
    while (num.back() == '0') {
      num.pop_back();
    }
    return num;
  }
};

func removeTrailingZeros(num string) string {
  i := len(num) - 1
  for num[i] == '0' {
    i--
  }
  return num[:i+1]
}

function removeTrailingZeros(num: string): string {
  let i = num.length - 1;
  while (num[i] === '0') {
    --i;
  }
  return num.substring(0, i + 1);
}

impl Solution {
  pub fn remove_trailing_zeros(num: String) -> String {
    let mut i = num.len() - 1;

    while num.chars().nth(i) == Some('0') {
      i -= 1;
    }

    num[..i + 1].to_string()
  }
}

Solution 2

impl Solution {
  pub fn remove_trailing_zeros(num: String) -> String {
    num.chars()
      .rev()
      .skip_while(|&c| c == '0')
      .collect::<String>()
      .chars()
      .rev()
      .collect::<String>()
  }
}