zoukankan      html  css  js  c++  java
  • [LeetCode] Palindrome Permutation I & II

    Palindrome Permutation

    Given a string, determine if a permutation of the string could form a palindrome.

    For example,
    "code" -> False, "aab" -> True, "carerac" -> True.

    Hint:

      1. Consider the palindromes of odd vs even length. What difference do you notice?
      2. Count the frequency of each character.
      3. If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times
     1 class Solution {
     2 public:
     3     bool canPermutePalindrome(string s) {
     4         vector<int> cnt(256, 0);
     5         for (auto a : s) ++cnt[a];
     6         bool flag = false;
     7         for (auto n : cnt) if (n & 1) {
     8             if (!flag) flag = true;
     9             else return false;
    10         }
    11         return true;
    12     }
    13 };

    Palindrome Permutation II

    Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

    For example:

    Given s = "aabb", return ["abba", "baab"].

    Given s = "abc", return [].

    Hint:

    1. If a palindromic permutation exists, we just need to generate the first half of the string.
    2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

    没按提示来,直接用的DFS,不知道符不符合要求。

     1 class Solution {
     2 public:
     3     void dfs(vector<string> &res, vector<int> &cnt, string &s, int l, int r) {
     4         if (l >= r) {
     5             res.push_back(s);
     6             return;
     7         }
     8         for (int i = 0; i < cnt.size(); ++i) if (cnt[i] >= 2) {
     9             cnt[i] -= 2;
    10             s[l] = s[r] = i;
    11             dfs(res, cnt, s, l + 1, r - 1);
    12             cnt[i] += 2;
    13         }
    14     }
    15     vector<string> generatePalindromes(string s) {
    16         vector<int> cnt(256, 0);
    17         for (auto a : s) ++cnt[a];
    18         bool flag = false;
    19         for (int i = 0; i < cnt.size(); ++i) if (cnt[i] & 1) {
    20             if (!flag) {
    21                 flag = true;
    22                 s[s.length() / 2] = i;
    23             } else {
    24                 return {};
    25             }
    26         }
    27         vector<string> res;
    28         dfs(res, cnt, s, 0, s.length() - 1);
    29         return res;
    30     }
    31 };
  • 相关阅读:
    F
    E
    网上见到一同行发的隐私政策 备以后用
    Cannot connect to the Docker daemon. Is the docker daemon running on this host?
    mark
    转 随机数问题
    随机不同的数
    转 基于Quick-cocos2dx 2.2.3 的动态更新实现完整篇。(打包,服务器接口,模块自更新
    字符串
    关于cmbiling.jar cocos2dx的问题
  • 原文地址:https://www.cnblogs.com/easonliu/p/4784930.html
Copyright © 2011-2022 走看看