我猜这个问题类似于 LeetCode 567.这些是简单、高效、低复杂性的公认解决方案:

I guess the question is similar to LeetCode 567. These are simple, efficient, low-complexity accepted solutions:

C#

class Solution {
    public bool CheckInclusion(string s1, string s2) {
        int lengthS1 = s1.Length;
        int lengthS2 = s2.Length;

        if (lengthS1 > lengthS2)
            return false;

        int[] countmap = new int[26];

        for (int i = 0; i < lengthS1; i++)
            countmap[s1[i] - 97]++;


        int[] bCount = new int[26];

        for (int i = 0; i < lengthS2; i++) {
            bCount[s2[i] - 97]++;

            if (i >= (lengthS1 - 1)) {
                if (allZero(countmap, bCount))
                    return true;

                bCount[s2[i - (lengthS1 - 1)] - 97]--;
            }
        }

        return false;
    }

    private bool allZero(int[] s1, int[] s2) {
        for (int i = 0; i < 26; i++) {
            if (s1[i] != s2[i])
                return false;
        }

        return true;
    }
}

Java

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int lengthS1 = s1.length();
        int lengthS2 = s2.length();

        if (lengthS1 > lengthS2)
            return false;

        int[] countmap = new int[26];

        for (int i = 0; i < lengthS1; i++) {
            countmap[s1.charAt(i) - 97]++;
            countmap[s2.charAt(i) - 97]--;
        }

        if (allZero(countmap))
            return true;

        for (int i = lengthS1; i < lengthS2; i++) {
            countmap[s2.charAt(i) - 97]--;
            countmap[s2.charAt(i - lengthS1) - 97]++;

            if (allZero(countmap))
                return true;
        }

        return false;
    }

    private boolean allZero(int[] count) {
        for (int i = 0; i < 26; i++)
            if (count[i] != 0)
                return false;

        return true;
    }
}

Python

class Solution:
    def checkInclusion(self, s1, s2):
        count_map_s1 = collections.Counter(s1)
        count_map_s2 = collections.Counter(s2[:len(s1)])

        for i in range(len(s1), len(s2)):
            if count_map_s1 == count_map_s2:
                return True

            count_map_s2[s2[i]] += 1
            count_map_s2[s2[i - len(s1)]] -= 1
            if count_map_s2[s2[i - len(s1)]] == 0:
                del(count_map_s2[s2[i - len(s1)]])

        return count_map_s2 == count_map_a

参考

代码在以下链接中解释:

Reference

The codes are explained in the following links:

• LeetCode 567 解决方案

LeetCode 567讨论区

这两个答案也很有用:

• 链接 1
• 链接 2