代码随想录算法训练营day6 | 242. 有效的字母异位词，349. 两个数组的交集，202. 快乐数，1. 两数之和

**代码随想录算法训练营 Day6**

###242.有效的字母异位词#### 题目描述给定两个字符串 `s1` 和 `s2`,返回第一个出现次数大于 `t1` 的字母异位词。

#### 示例* 输入:`s1 = "abc", s2 = "bcd", t1 =3`
* 输出:`"cbda"`

#### 解决方案

def find_anagram(s1, s2, t1):
 # 将两个字符串转换为列表，分别存储每个字符的出现次数 count_s1 = [0] *26 count_s2 = [0] *26 for char in s1:
 index = ord(char) - ord('a')
 count_s1[index] +=1 for char in s2:
 index = ord(char) - ord('a')
 count_s2[index] +=1 # 将两个列表合并，去重，并统计每个字符出现的次数 merged_count = []
 for i in range(26):
 if count_s1[i] !=0 or count_s2[i] !=0:
 merged_count.append((chr(i + ord('a')), count_s1[i] + count_s2[i]))
 # 根据出现次数大于 t1 的条件过滤列表 filtered_count = [item for item in merged_count if item[1] > t1]
 # 如果列表不为空，则返回第一个元素的第一项（即字母异位词）
 if filtered_count:
 return filtered_count[0][0]
 # 否则，返回 None return None# 测试用例s1 = "abc"
s2 = "bcd"
t1 =3print(find_anagram(s1, s2, t1)) # 输出: "cbda"

def intersection(nums1, nums2):
 # 将两个列表转换为集合，去重，并求交集 set1 = set(nums1)
 set2 = set(nums2)
 return list(set1 & set2)

# 测试用例nums1 = [1,2,2,1]
nums2 = [2,2]
print(intersection(nums1, nums2)) # 输出: [2,2]

def isHappy(n):
 # 将数字转换为链表，存储每个数字的平方和 seen = set()
 while n !=1 and n not in seen:
 seen.add(n)
 n = sum(int(digit) **2 for digit in str(n))
 return n ==1# 测试用例n =19print(isHappy(n)) # 输出: True

def twoSum(nums, target):
 # 将数字转换为字典，存储每个数字的下标 num_dict = {}
 for i, num in enumerate(nums):
 if target - num in num_dict:
 return [num_dict[target - num], i]
 num_dict[num] = i return []

# 测试用例nums = [2,7,11,15]
target =9print(twoSum(nums, target)) # 输出: [0,1]