From 3caaf866f4e2ec8ce846b261dbf5b07b1a828aa4 Mon Sep 17 00:00:00 2001 From: ironartisan Date: Sat, 7 Aug 2021 10:25:15 +0800 Subject: [PATCH 1/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201002.=E6=9F=A5?= =?UTF-8?q?=E6=89=BE=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6python3=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 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) { From 660833125d7e7d7e44da1d68deb53430640f6bc7 Mon Sep 17 00:00:00 2001 From: ironartisan Date: Sat, 7 Aug 2021 21:15:06 +0800 Subject: [PATCH 2/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201207.=E7=8B=AC?= =?UTF-8?q?=E4=B8=80=E6=97=A0=E4=BA=8C=E7=9A=84=E5=87=BA=E7=8E=B0=E6=AC=A1?= =?UTF-8?q?=E6=95=B0python3=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1207.独一无二的出现次数.md | 16 +++++++++++++++- 1 file changed, 15 insertions(+), 1 deletion(-) diff --git a/problems/1207.独一无二的出现次数.md b/problems/1207.独一无二的出现次数.md index c1720430..32aa3c3a 100644 --- a/problems/1207.独一无二的出现次数.md +++ b/problems/1207.独一无二的出现次数.md @@ -100,7 +100,21 @@ class Solution { ``` Python： - +```python +class Solution: + def uniqueOccurrences(self, arr: List[int]) -> bool: + count = [0] * 2002 + for i in range(len(arr)): + count[arr[i] + 1000] += 1 # 防止负数作为下标 + freq = [False] * 1002 # 标记相同频率是否重复出现 + for i in range(2001): + if count[i] > 0: + if freq[count[i]] == False: + freq[count[i]] = True + else: + return False + return True +``` Go： JavaScript： From 0076ddab1998c3846430eb436ebfc7e4ff23a075 Mon Sep 17 00:00:00 2001 From: ironartisan Date: Sun, 8 Aug 2021 08:54:35 +0800 Subject: [PATCH 3/3] =?UTF-8?q?=E4=BF=AE=E6=AD=A30349.=E4=B8=A4=E4=B8=AA?= =?UTF-8?q?=E6=95=B0=E7=BB=84=E7=9A=84=E4=BA=A4=E9=9B=86python3=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0349.两个数组的交集.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 5c635d39..5dc18478 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -127,7 +127,7 @@ class Solution: for num in nums2: if num in set1: result_set.add(num) # set1里出现的nums2元素 存放到结果 - return result_set + return list(result_set) ```