CC189 - 1.4

最新推荐文章于 2021-08-01 21:20:46 发布

real_lisa

最新推荐文章于 2021-08-01 21:20:46 发布

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分类专栏： CC189

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博客围绕判断字符串是否为回文排列展开。指出回文排列中大多数字符数量为偶数，奇数长度字符串最多一个字符数量为奇数。还给出三种解法，包括两次迭代、一次迭代中检查以及使用位向量，并提供了相应代码链接。

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1.4 Palindrome Permutation: Given a string, write a function to check if it is a permutation of a palindrome. A palindrome is a word or phrase that is the same forwards and backwards. A permutation is a rearrangement of letters. The palindrome does not need to be limited to just dictionary words.

Highlight:
How to better interpret a permutation of a palindrome?
The word has an even number of almost all characters. At most one character in the middle can have a odd count.
-> even string length: all even counts
-> odd string length: at most one odd count

解法一：two iterations

bool isPalindromePurmutation(string str){
    unordered_map<char, int> m;
    for(char c: str){
        m[c]++;
    }
    bool hasOdd = str.size()%2;
    for(auto it=m.begin(); it!=m.end(); it++){
        int count = it->second;
        if(count%2!=0){
            if(hasOdd) {
                hasOdd=0;
            }else{
                return false;
            }
        }
    }
    return true;
}

https://onlinegdb.com/rJyPYVaiV

解法二：one iteration - checking along the iteration

bool isPalindromePurmutation(string str){
    unordered_map<char, int> m;
    int countOdd = 0;
    for(char c: str){
        m[c]++;
        if(m[c]%2==1){
            countOdd++;
        }else{
            countOdd--;
        }
    }
    return countOdd<=1;
}

https://onlinegdb.com/SkL3hETi4

解法三：use a bit vector

There is an elegant way to check if only one bit turns to 1:
00010000 - 1 = 00001111
00010000 & 00001111 = 0

bool isPalindromePurmutation(string str){
    int bit = 0;
    for(char c: str){
        int pos = c-'a';
        bit ^= 1<<pos;
    }
    if(((bit-1)&bit)==0){
        return true;
    }
    return false;
}

https://onlinegdb.com/B1R8kLTjV