剖析 gh_mirrors/leet 项目：字符串重排题解的算法优化逻辑（131）

# 🔍 剖析 gh_mirrors/leet 项目：字符串重排题解的算法优化逻辑 🚀

在 gh_mirrors/leet 项目中，字符串重排问题是一个经典算法挑战，其核心在于高效判断两个字符串是否互为排列组合。让我们深入解析其优化逻辑！💡

## 🔧 基础解法分析
最直观的解法是**排序比较法**：将两个字符串排序后直接比较。虽然简单易懂，但时间复杂度为 O(n log n)，不够高效。😕

```python
def isAnagram(s: str, t: str) -> bool:
  return sorted(s) == sorted(t)
```

## ⚡ 哈希表优化
更聪明的做法是使用**哈希表计数法**，将时间复杂度降至 O(n)！🎯

1. **长度快速判断**：先比较字符串长度，不同则直接返回False
2. **字符频率统计**：使用字典统计每个字符出现次数
3. **双哈希表比较**：比较两个字符串的字符频率是否完全一致

```python
from collections import defaultdict

def isAnagram(s: str, t: str) -> bool:
  if len(s) != len(t):
    return False
  
  count_s = defaultdict(int)
  count_t = defaultdict(int)
  
  for char in s:
    count_s[char] += 1
  
  for char in t:
    count_t[char] += 1
  
  return count_s == count_t
```

## 🏆 终极优化：单哈希表
可以进一步优化空间复杂度，只使用一个哈希表：

1. 先增加s字符串的字符计数
2. 再减少t字符串的字符计数
3. 最后检查所有计数是否归零

```python
def isAnagram(s: str, t: str) -> bool:
  if len(s) != len(t):
    return False
  
  counter = [0] * 26 # 假设只包含小写字母
  
  for char in s:
    counter[ord(char) - ord('a')] += 1
  
  for char in t:
    counter[ord(char) - ord('a')] -= 1
    if counter[ord(char) - ord('a')] < 0:
      return False
  
  return True
```

## 🌟 总结
从 O(n log n) 到 O(n) 的优化过程展示了算法思维的魅力！选择合适的数据结构（哈希表）和巧妙的计数方法，能显著提升性能。在实际面试中，建议从基础解法开始，逐步优化，并解释每步优化的考量。🧠✨
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