567. 字符串的排列

题目链接：https://leetcode-cn.com/problems/permutation-in-string/
执行用时：44 ms, 在所有 Swift 提交中击败了84.00%的用户
内存消耗：14.2 MB, 在所有 Swift 提交中击败了38.00%的用户

class Solution {
    func checkInclusion(_ s1: String, _ s2: String) -> Bool {
        if s1.count > s2.count {
            return false
        }
        let a = Array(s1), b = Array(s2), m = s1.count, n = s2.count
        var count1 = [Int](repeating: 0, count: 26)
        var count2 = [Int](repeating: 0, count: 26)
        //比较短字符串长度下是否是对应排列的子串，检查元素
        for i in 0 ..< m {
            count1[Int(a[i].asciiValue!) - 97] += 1
            count2[Int(b[i].asciiValue!) - 97] += 1
        }
        if count1.elementsEqual(count2) {
            return true
        }
        for i in m ..< n {
            //滑动窗口向后移动，去掉滑动窗口前一个元素
            count2[Int(b[i].asciiValue!) - 97] += 1
            count2[Int(b[i-m].asciiValue!) - 97] -= 1
            if count1.elementsEqual(count2) {
                return true
            }
        }
        return false
    }
}