Question

1881. Maximum Value after Insertion

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Description

You are given a very large integer n, represented as a string,​​​​​​ and an integer digit x. The digits in n and the digit x are in the inclusive range [1, 9], and n may represent a negative number.

You want to maximize n's numerical value by inserting x anywhere in the decimal representation of n​​​​​​. You cannot insert x to the left of the negative sign.

• For example, if n = 73 and x = 6, it would be best to insert it between 7 and 3, making n = 763.
• If n = -55 and x = 2, it would be best to insert it before the first 5, making n = -255.

Return _a string representing the maximum value of _n_​​​​​​ after the insertion_.

 

Example 1:

Input: n = "99", x = 9
Output: "999"
Explanation: The result is the same regardless of where you insert 9.

Example 2:

Input: n = "-13", x = 2
Output: "-123"
Explanation: You can make n one of {-213, -123, -132}, and the largest of those three is -123.

 

Constraints:

• 1 <= n.length <= 105
• 1 <= x <= 9
• The digits in n​​​ are in the range [1, 9].
• n is a valid representation of an integer.
• In the case of a negative n,​​​​​​ it will begin with '-'.

Solutions

Solution 1

class Solution:
  def maxValue(self, n: str, x: int) -> str:
    if n[0] != '-':
      for i, c in enumerate(n):
        if int(c) < x:
          return n[:i] + str(x) + n[i:]
      return n + str(x)
    else:
      for i, c in enumerate(n[1:]):
        if int(c) > x:
          return n[: i + 1] + str(x) + n[i + 1 :]
      return n + str(x)

class Solution {
  public String maxValue(String n, int x) {
    int i = 0;
    if (n.charAt(0) != '-') {
      for (; i < n.length() && n.charAt(i) - '0' >= x; ++i)
        ;
    } else {
      for (i = 1; i < n.length() && n.charAt(i) - '0' <= x; ++i)
        ;
    }
    return n.substring(0, i) + x + n.substring(i);
  }
}

class Solution {
public:
  string maxValue(string n, int x) {
    int i = 0;
    if (n[0] != '-')
      for (; i < n.size() && n[i] - '0' >= x; ++i)
        ;
    else
      for (i = 1; i < n.size() && n[i] - '0' <= x; ++i)
        ;
    return n.substr(0, i) + to_string(x) + n.substr(i);
  }
};

func maxValue(n string, x int) string {
  i := 0
  y := byte('0' + x)
  if n[0] != '-' {
    for ; i < len(n) && n[i] >= y; i++ {
    }
  } else {
    for i = 1; i < len(n) && n[i] <= y; i++ {
    }
  }
  return n[:i] + string(y) + n[i:]
}

/**
 * @param {string} n
 * @param {number} x
 * @return {string}
 */
var maxValue = function (n, x) {
  let nums = [...n];
  let sign = 1,
    i = 0;
  if (nums[0] == '-') {
    sign = -1;
    i++;
  }
  while (i < n.length && (nums[i] - x) * sign >= 0) {
    i++;
  }
  nums.splice(i, 0, x);
  return nums.join('');
};