438. Find All Anagrams in a String

 

Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

 

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

 

这道题给了我们两个字符串s和p,让在s中找字符串p的所有变位次的位置,所谓变位次就是字符种类个数均相同但是顺序可以不同的两个词,那么肯定首先就要统计字符串p中字符出现的次数,然后从s的开头开始,每次找p字符串长度个字符,来验证字符个数是否相同,如果不相同出现了直接 break,如果一直都相同了,则将起始位置加入结果 res 中,参见代码如下:

 

解法一:

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        if (s.empty()) return {};
        vector<int> res, cnt(128, 0);
        int ns = s.size(), np = p.size(), i = 0;
        for (char c : p) ++cnt[c];
        while (i < ns) {
            bool success = true;
            vector<int> tmp = cnt;
            for (int j = i; j < i + np; ++j) {
                if (--tmp[s[j]] < 0) {
                    success = false;
                    break;
                }
            }
            if (success) {
                res.push_back(i); 
            }
            ++i;
        }
        return res;
    }
};

 

我们可以将上述代码写的更加简洁一些,用两个哈希表,分别记录p的字符个数,和s中前p字符串长度的字符个数,然后比较,如果两者相同,则将0加入结果 res 中,然后开始遍历s中剩余的字符,每次右边加入一个新的字符,然后去掉左边的一个旧的字符,每次再比较两个哈希表是否相同即可,参见代码如下:

 

解法二:

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        if (s.empty()) return {};
        vector<int> res, m1(256, 0), m2(256, 0);
        for (int i = 0; i < p.size(); ++i) {
            ++m1[s[i]]; ++m2[p[i]];
        }
        if (m1 == m2) res.push_back(0);
        for (int i = p.size(); i < s.size(); ++i) {
            ++m1[s[i]]; 
            --m1[s[i - p.size()]];
            if (m1 == m2) res.push_back(i - p.size() + 1);
        }
        return res;
    }
};

 

下面这种利用滑动窗口 Sliding Window 的方法也比较巧妙,首先统计字符串p的字符个数,然后用两个变量 left 和 right 表示滑动窗口的左右边界,用变量 cnt 表示字符串p中需要匹配的字符个数,然后开始循环,如果右边界的字符已经在哈希表中了,说明该字符在p中有出现,则 cnt 自减1,然后哈希表中该字符个数自减1,右边界自加1,如果此时 cnt 减为0了,说明p中的字符都匹配上了,那么将此时左边界加入结果 res 中。如果此时 right 和 left 的差为p的长度,说明此时应该去掉最左边的一个字符,如果该字符在哈希表中的个数大于等于0,说明该字符是p中的字符,为啥呢,因为上面有让每个字符自减1,如果不是p中的字符,那么在哈希表中个数应该为0,自减1后就为 -1,所以这样就知道该字符是否属于p,如果去掉了属于p的一个字符,cnt 自增1,参见代码如下:

 

解法三:

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        if (s.empty()) return {};
        vector<int> res, m(256, 0);
        int left = 0, right = 0, cnt = p.size(), n = s.size();
        for (char c : p) ++m[c];
        while (right < n) {
            if (m[s[right++]]-- >= 1) --cnt;
            if (cnt == 0) res.push_back(left);
            if (right - left == p.size() && m[s[left++]]++ >= 0) ++cnt;
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/438

 

类似题目:

Valid Anagram

Anagrams

 

参考资料:

https://leetcode.com/problems/find-all-anagrams-in-a-string/

https://leetcode.com/problems/find-all-anagrams-in-a-string/discuss/92015/ShortestConcise-JAVA-O(n)-Sliding-Window-Solution

https://leetcode.com/problems/find-all-anagrams-in-a-string/discuss/92027/c-on-sliding-window-concise-solution-with-explanation

https://leetcode.com/problems/find-all-anagrams-in-a-string/discuss/92007/Sliding-Window-algorithm-template-to-solve-all-the-Leetcode-substring-search-problem.

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation