有效的字母异位词 - 简单-优快云博客

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C++

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Have a quick eye on the topic.

The first instinct is sort algorithm. Sort is a standard library in c++. The basic usage of sort is as folllow.

vector<int> nums = {5, 2, 9, 1, 5, 6};

// 对 vector 排序
sort(nums.begin(), nums.end());

输出：1， 2， 5， 5， 6， 9


// 使用自定义比较函数实现降序排序
sort(nums.begin(), nums.end(), greater<int>());
输出： 9， 6， 5， 5， 2， 1

and it really works.

class Solution {
public:
    bool isAnagram(string s, string t) {

        // 对两个字符串进行排序
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());

        // 比较排序后的字符串是否相等
        return s == t;
    }
};

But it something llike a trick. Snbsequence is the key point of this type of topics. Count the numbers of each letters in two strings. If there are 13 'a' in string s while 38 'a' in string t, then return false.

So for people can understand is count[a] = 13, which means letter a appears 13 times. But computer donot think so. Memory is everything to the computer. Computer knows 0 and 1. So count [1] = 13 means the object in storage position 1 shows up 13 times.

‘a’ corresponds to index 0
‘b’ corresponds to index 1
...
‘z’ corresponds to index 25

so make a count array, 26 elements in the array is 0 at first.

int count(26) = {0};

which meas 26 elements in the count array is 0;

position	0	1	2	3	...	25
corespond letter	a	b	c	d		z
value	0	0	0	0	0	0

counts[s[i] - 'a']++; // 对s中的字符加1

Suppose the string s = "hello"

字符	`c - 'a'` 的值
`'h'`	7
`'e'`	4
`'l'`	11
`'l'`	11
`'o'`	14

Then the code can be writen.

class Solution {
public:
    bool isAnagram(string s, string t) {

        int counts[26] = {0}; // 初始化计数数组
        int countt[26] = {0};

        // 统计字符出现次数
        for (char c : s) counts[c - 'a']++;
        for (char c : t) countt[c - 'a']++;

        // 比较
        return equal(begin(counts), end(counts), begin(countt));
    }
};

The type of the variables are changed from int to vector. It wii make changes.

// 初始化一个有13个数字的向量，全部为0
vector<int> diyname(13, 0)

class Solution {
public:
    bool isAnagram(string s, string t) {

// int变成 Vector
        vector<int> counts(26, 0); // 初始化计数数组
        vector<int> countt(26, 0);

        // 统计字符出现次数
        for (char c : s) counts[c - 'a']++;
        for (char c : t) countt[c - 'a']++;

        // 比较
        return counts == countt;
    }
};

But both the two methods initialize two arrays. What if just one array have to initialize?

class Solution {
public:
    bool isAnagram(string s, string t) {

        int count[26] = {0}; // 初始化计数数组

        // 统计字符出现次数
        for (char c : s) count[c - 'a']++;
        for (char c : t) count[c - 'a']--;

        // 检查所有字符的计数是否为零
        for (int i = 0; i < 26; i++) 
        {
            if (count[i] != 0) 
            {
                return false;
            }
        }

        return true;
    }
};

Methods are flexible. Follow the rules in c++ of each gramma and the code works. At the very begining, I confused about the gramma. Now I donot. Stepping in the new land always come up with strangers. After some practice, the strangers become neighbours.