LeetCode每日一题(1881. Maximum Value after Insertion)

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 ‘-’.

---

刨除正负号只看数字部分， 如果我们想让数字尽可能的大， 我们需要从高位到地位找到第一个< x 的数字， 将 x 插在该数字之前， 相反，如果我们想让数字尽可能的小， 我们需要从高位到低位找到第一个 > x 的数字， 将 x 插在该数字之前。简单来说就是插入的位数越高对整体数字大小的变化影响越大， 这个是量， 如果插入的 x 大于原来在此位置上的数字，那数字整体会变大， 反之则变小, 这个是质。

---

use std::collections::HashMap;

impl Solution {
    pub fn max_value(mut n: String, x: i32) -> String {
        let m: HashMap<char, i32> = vec![
            ('1', 1),
            ('2', 2),
            ('3', 3),
            ('4', 4),
            ('5', 5),
            ('6', 6),
            ('7', 7),
            ('8', 8),
            ('9', 9),
        ]
        .into_iter()
        .collect();
        if n.starts_with('-') {
            for (i, c) in n.chars().enumerate().skip(1) {
                let v = *m.get(&c).unwrap();
                if v > x {
                    n.insert_str(i, &x.to_string());
                    return n;
                }
            }
            n.push_str(&x.to_string());
            return n;
        }
        for (i, c) in n.chars().enumerate() {
            let v = *m.get(&c).unwrap();
            if v < x {
                n.insert_str(i, &x.to_string());
                return n;
            }
        }
        n.push_str(&x.to_string());
        return n;
    }
}