Question

1323. Maximum 69 Number

中文文档

Description

You are given a positive integer num consisting only of digits 6 and 9.

Return _the maximum number you can get by changing at most one digit (_6_ becomes _9_, and _9_ becomes _6_)_.

 

Example 1:

Input: num = 9669
Output: 9969
Explanation: 
Changing the first digit results in 6669.
Changing the second digit results in 9969.
Changing the third digit results in 9699.
Changing the fourth digit results in 9666.
The maximum number is 9969.

Example 2:

Input: num = 9996
Output: 9999
Explanation: Changing the last digit 6 to 9 results in the maximum number.

Example 3:

Input: num = 9999
Output: 9999
Explanation: It is better not to apply any change.

 

Constraints:

• 1 <= num <= 104
• num consists of only 6 and 9 digits.

Solutions

Solution 1

class Solution:
  def maximum69Number(self, num: int) -> int:
    return int(str(num).replace("6", "9", 1))

class Solution {
  public int maximum69Number(int num) {
    return Integer.valueOf(String.valueOf(num).replaceFirst("6", "9"));
  }
}

class Solution {
public:
  int maximum69Number(int num) {
    string s = to_string(num);
    for (char& ch : s) {
      if (ch == '6') {
        ch = '9';
        break;
      }
    }
    return stoi(s);
  }
};

func maximum69Number(num int) int {
  s := strconv.Itoa(num)
  nums := []byte(s)
  for i, ch := range nums {
    if ch == '6' {
      nums[i] = '9'
      break
    }
  }
  ans, _ := strconv.Atoi(string(nums))
  return ans
}

function maximum69Number(num: number): number {
  return Number((num + '').replace('6', '9'));
}

impl Solution {
  pub fn maximum69_number(num: i32) -> i32 {
    num.to_string().replacen('6', "9", 1).parse().unwrap()
  }
}

class Solution {
  /**
   * @param Integer $num
   * @return Integer
   */
  function maximum69Number($num) {
    $num = strval($num);
    $n = strpos($num, '6');
    $num[$n] = 9;
    return intval($num);
  }
}

int maximum69Number(int num) {
  int n = 0;
  int i = 0;
  int t = num;
  while (t) {
    n++;
    if (t % 10 == 6) {
      i = n;
    }
    t /= 10;
  }
  return num + 3 * pow(10, i - 1);
}