567. Permutation in String

解題說明

C++ 解法

複雜度分析

虛擬碼

1. If len(s1) > len(s2): return false
2. Build count1[26] for s1, count2[26] for first window of s2 (size = len(s1))
3. Count initial matches = number of indices i where count1[i] == count2[i]
4. Set l = 0
5. For r from len(s1) to len(s2) - 1:
   a. If matches == 26: return true
   b. Add s2[r] to window: increment count2[s2[r]], update matches
   c. Remove s2[l] from window: decrement count2[s2[l]], update matches
   d. l++
6. Return matches == 26

其他解法

延伸思考