求两个字符串的含重复项公共元素（需匹配重复次数）

开发者阿泽

2026-8-25

解决方法

最简实现思路是先分别统计两个字符串中每个字符的出现次数，对每个共同字符取两个次数的较小值，再按该次数重复字符并拼接成结果，全程无需使用集合交集操作。

方案一：利用标准库`Counter`（简洁高效）

from collections import Counter

s1 = 'aebcdee'
s2 = 'aaeedfskm'

# 统计两个字符串的字符出现次数
count_s1 = Counter(s1)
count_s2 = Counter(s2)

common_list = []
# 遍历第一个字符串的字符统计结果
for char, cnt in count_s1.items():
    # 仅处理第二个字符串中存在的字符
    if char in count_s2:
        # 取两个次数的最小值，重复字符后加入列表
        common_list.append(char * min(cnt, count_s2[char]))

# 拼接成最终结果字符串
common = ''.join(common_list)
print(common)  # 输出: aeed

方案二：手动统计字符次数（无需导入库）

如果不想依赖标准库，也可以手动实现字符计数逻辑：

def count_char_frequency(s):
    freq = {}
    for char in s:
        freq[char] = freq.get(char, 0) + 1
    return freq

s1 = 'aebcdee'
s2 = 'aaeedfskm'

freq1 = count_char_frequency(s1)
freq2 = count_char_frequency(s2)

common_list = []
for char in freq1:
    if char in freq2:
        common_list.append(char * min(freq1[char], freq2[char]))

common = ''.join(common_list)
print(common)  # 输出: aeed