README.md

有效的字母异位词

难度：简单

https://leetcode.cn/problems/valid-anagram

题目

给定两个字符串 s 和 t ，编写一个函数来判断 t 是否是 s 的字母异位词。

注意：若 s 和 t 中每个字符出现的次数都相同，则称 s 和 t 互为字母异位词。

示例 1:

输入: s = "anagram", t = "nagaram"
输出: true

示例 2:

输入: s = "rat", t = "car"
输出: false

题解

TS

export function isAnagram(s: string, t: string): boolean {
  if (s.length !== t.length) {
    return false
  }
  const mapS = new Map<string, number>();
  const mapT = new Map<string, number>();

  for (let i = 0; i < s.length; i++) {
    mapS.set(s[i], (mapS.get(s[i]) || 0) + 1);
    mapT.set(t[i], (mapT.get(t[i]) || 0) + 1);
  }

  if (mapS.size !== mapT.size) {
    return false;
  }

  let ans = true;

  for (const key of mapS.keys()) {
    if (mapS.get(key) !== mapT.get(key)) {
      ans = false;
      break;
    }
  }
  return ans
}

// copilot 生成
export function isAnagram1(s: string, t: string): boolean {
  if (s.length !== t.length) {
    return false;
  }
  const mapS = new Map<string, number>();
  const mapT = new Map<string, number>();

  // Count the frequency of each character in string s
  for (const char of s) {
    mapS.set(char, (mapS.get(char) || 0) + 1);
  }
  for (const char of t) {
    mapT.set(char, (mapT.get(char) || 0) + 1);
  }

  // Count the frequency of each character in string t
  for (const [char, count] of mapS) {
    if (mapT.get(char) !== count) {
      return false;
    }
  }

  return true;
}

Python

class Solution:
  def isAnagram(self, s: str, t: str) -> bool:
    # 判断长度
    if len(s) != len(t): 
      return False
    
    mapS: dict = {}
    mapT: dict = {}
    for i in range(len(s)):
      if s[i] in mapS:
        mapS[s[i]] = mapS[s[i]] + 1
      else:
        mapS[s[i]] = 1

      if t[i] in mapT:
        mapT[t[i]] = mapT[t[i]] + 1
      else:
        mapT[t[i]] = 1

    if len(mapS) != len(mapT):
      return False
  
    for key in mapS.keys():
      if not (key in mapT) or (mapS[key] != mapT[key]):
        return False
      # if mapS[key] != mapT[key]:
      #   return False
      # if (key in mapT) and (mapS[key] != mapT[key]):
      #   return False
    return True


s = Solution()
# print(s.isAnagram('anagram','nagaram'))
print(s.isAnagram('rat','car'))

# a: dict = {}
# a['a'] = 1
# a['b'] = 2
# a['c'] = 3

# print(a['a'])