From 3caaf866f4e2ec8ce846b261dbf5b07b1a828aa4 Mon Sep 17 00:00:00 2001
From: ironartisan <ironerhalo@gmail.com>
Date: Sat, 7 Aug 2021 10:25:15 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201002.=E6=9F=A5=E6=89=BE?=
 =?UTF-8?q?=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6python3=E7=89=88=E6=9C=AC?=
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---
 problems/1002.查找常用字符.md | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)

diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md
index e3014d00..19f9e128 100644
--- a/problems/1002.查找常用字符.md
+++ b/problems/1002.查找常用字符.md
@@ -169,6 +169,29 @@ class Solution {
     }
 }
 ```
+```python
+class Solution:
+    def commonChars(self, words: List[str]) -> List[str]:
+        if not words: return []
+        result = []
+        hash = [0] * 26 # 用来统计所有字符串里字符出现的最小频率
+        for i, c in enumerate(words[0]):  # 用第一个字符串给hash初始化
+            hash[ord(c) - ord('a')] += 1
+        # 统计除第一个字符串外字符的出现频率
+        for i in range(1, len(words)):
+            hashOtherStr = [0] * 26
+            for j in range(len(words[0])):
+                hashOtherStr[ord(words[i][j]) - ord('a')] += 1
+            # 更新hash，保证hash里统计26个字符在所有字符串里出现的最小次数
+            for k in range(26):
+                hash[k] = min(hash[k], hashOtherStr[k])
+        # 将hash统计的字符次数，转成输出形式
+        for i in range(26):
+            while hash[i] != 0: # 注意这里是while，多个重复的字符
+                result.extend(chr(i + ord('a')))
+                hash[i] -= 1
+        return result
+```
 javaScript
 ```js
 var commonChars = function (words) {