[LeetCode] 1370. Increasing Decreasing String
You are given a string s
. Reorder 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.
Constraints:
1 <= s.length <= 500
s
consists of only lowercase English letters.
上升下降字符串。
给你一个字符串 s ,请你根据下面的算法重新构造字符串:
从 s 中选出 最小 的字符,将它 接在 结果字符串的后面。
从 s 剩余字符中选出 最小 的字符,且该字符比上一个添加的字符大,将它 接在 结果字符串后面。
重复步骤 2 ,直到你没法从 s 中选择字符。
从 s 中选出 最大 的字符,将它 接在 结果字符串的后面。
从 s 剩余字符中选出 最大 的字符,且该字符比上一个添加的字符小,将它 接在 结果字符串后面。
重复步骤 5 ,直到你没法从 s 中选择字符。
重复步骤 1 到 6 ,直到 s 中所有字符都已经被选过。
在任何一步中,如果最小或者最大字符不止一个 ,你可以选择其中任意一个,并将其添加到结果字符串。请你返回将 s 中字符重新排序后的 结果字符串 。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/increasing-decreasing-string
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题意还算直观,把input字符串重新排列。排列的规则是按字典序先从小到大,然后再从字典序最大的字母开始再往字典序最小的字母走。思路是用一个freq数组统计一下26个小写字母的出现次数,然后按照规则把字母重新拼接成字符串即可。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public String sortString(String s) { 3 int len = s.length(); 4 int[] freq = new int[26]; 5 for (int i = 0; i < len; i++) { 6 freq[s.charAt(i) - 'a']++; 7 } 8 StringBuilder sb = new StringBuilder(); 9 int count = 0; 10 while (count < len) { 11 for (int i = 0; i < 26; i++) { 12 if (freq[i] > 0) { 13 sb.append((char) (i + 'a')); 14 freq[i]--; 15 count++; 16 } 17 } 18 19 for (int i = 25; i >= 0; i--) { 20 if (freq[i] > 0) { 21 sb.append((char) (i + 'a')); 22 freq[i]--; 23 count++; 24 } 25 } 26 } 27 return sb.toString(); 28 } 29 }
JavaScript实现
1 /** 2 * @param {string} s 3 * @return {string} 4 */ 5 var sortString = function (s) { 6 let len = s.length; 7 let freq = new Array(26).fill(0); 8 for (let i = 0; i < len; i++) { 9 freq[s.charCodeAt(i) - 97]++; 10 } 11 12 let res = []; 13 let count = 0; 14 while (count < len) { 15 for (let i = 0; i < 26; i++) { 16 if (freq[i] > 0) { 17 res.push(String.fromCharCode(i + 97)); 18 freq[i]--; 19 count++; 20 } 21 } 22 23 for (let i = 25; i >= 0; i--) { 24 if (freq[i] > 0) { 25 res.push(String.fromCharCode(i + 97)); 26 freq[i]--; 27 count++; 28 } 29 } 30 } 31 return res.join(''); 32 };