【字符串】C045_LC_上升下降字符串（思維）

一、Problem

Given a string s. You should re-order the string using the following algorithm:

1. Pick the smallest character from s and append it to the result.
2. Pick the smallest character from s which is greater than the last appended character to the result and append it.
3. Repeat step 2 until you cannot pick more characters.
4. Pick the largest character from s and append it to the result.
5. Pick the largest character from s which is smaller than the last appended character to the result and append it.
6. Repeat step 5 until you cannot pick more characters.
7. Repeat the steps from 1 to 6 until you pick all characters from s.

In each step, If the smallest or the largest character appears more than once you can choose any occurrence and append it to the result.

Return the result string after sorting s with this algorithm.

Input: s = "aaaabbbbcccc"
Output: "abccbaabccba"
Explanation: After steps 1, 2 and 3 of the first iteration, result = "abc"
After steps 4, 5 and 6 of the first iteration, result = "abccba"
First iteration is done. Now s = "aabbcc" and we go back to step 1
After steps 1, 2 and 3 of the second iteration, result = "abccbaabc"
After steps 4, 5 and 6 of the second iteration, result = "abccbaabccba"

二、Solution

方法一：思維

• 第一眼感覺很複雜啊，想了一會，發現題目已經給出規律了。
• 先按照升序字典序添加字母 a…z 各一次，然後按照降序字典序逐一添加字母 z…a 各一次，當然前提是字母要存在。
• 這不就是兩個循環可以解決的事嗎…

class Solution {
    public String sortString(String s) {
        StringBuilder sb = new StringBuilder();
        char[] cs = s.toCharArray();
        int[] m = new int[26];
        for (char c : cs) 
        	m[c-'a']++;
        while (sb.length() < cs.length) {
            for (int i = 0; i < 26; i++) {
                if (m[i]-- > 0)
                    sb.append((char) (i+'a'));
            }
            for (int i = 25; i >= 0; i--) {
                if (m[i]-- > 0)
                    sb.append((char) (i+'a'));
            }
        }
        return sb.toString();
    }
}

複雜度分析

• 時間複雜度：$O(n)$，
• 空間複雜度：$O(n)$，