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.
A:
class Solution { public: string sortString(string s) { vector<int> V(26,0); vector<char> res; for(auto c:s){ V[c-'a']++; } while(res.size() < s.length()){ // repeat 1,2 for(int i =0;i<26;i++){ if(V[i]>0){ res.push_back('a'+i ); V[i]--; } } // repeat 4,5,6, for(int i =25;i>=0;i--){ if(V[i]>0){ res.push_back('a'+i ); V[i]--; } } } string ss(res.begin(), res.end()); return ss; } };
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