1370. Increasing Decreasing String*

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最新推荐文章于 2020-12-04 16:07:24 发布

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分类专栏： LeetCode 文章标签： leetcode 算法

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本文详细解析了LeetCode上一道关于字符串重新排序的算法题——Increasing Decreasing String，介绍了如何通过哈希表（或数组）来实现字符串中字符的计数，并运用特定的算法步骤进行排序，最终得到按先增后减顺序排列的新字符串。

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1370. Increasing Decreasing String*

https://leetcode.com/problems/increasing-decreasing-string/

题目描述

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

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

Example 1:

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"

Example 2:

Input: s = "rat"
Output: "art"
Explanation: The word "rat" becomes "art" after re-ordering it with the mentioned algorithm.

Example 3:

Input: s = "leetcode"
Output: "cdelotee"

Example 4:

Input: s = "ggggggg"
Output: "ggggggg"

Example 5:

Input: s = "spo"
Output: "ops"

Constraints:

1 <= s.length <= 500
s contains only lower-case English letters.

C++ 实现 1

参考 [Java/Python 3] Two clean codes w/ explanation and analysis.

但这里采用哈希表解决, 当然也可以使用大小为 26 的数组来表示哈希表, 用于对字符串 s 中字符的计数. 引入辅助函数 add, 如果 asc 为 true, 那么在哈希表中从小到大查找字符; 如果为 false, 那么从大到小查找字符.

class Solution {
private:
    void add(string &res, unordered_map<char, int> &record, bool asc) {
        for (int i = 0; i < 26; ++ i) {
            auto idx = asc ? i : 25 - i;
            char c = 'a' + idx;
            if (record.count(c)) {
                res += c;
                record[c] --;
                if (record[c] == 0) record.erase(c);
            }
        }
    }
public:
    string sortString(string s) {
        unordered_map<char, int> record;
        for (auto &c : s) record[c] ++;
        string res;
        while (res.size() < s.size()) {
            add(res, record, true);
            add(res, record, false);
        }
        return res;
    }
};