Merge branch 'main' of github.com:doocs/leetcode into main

Author: yanglbme

images/contributors.svg

@@ -6,57 +6,59 @@
 
 **示例:**
 
-````
+```
 s = "abaccdeff"
 返回 "b"
 
 s = ""
 返回 " "
-``` 
+```
 
 **限制：**
 
 - `0 <= s 的长度 <= 50000`
 
 ## 解法
-有序字典解决。
+
+对字符串进行两次遍历：
+
+第一遍，先用 hash 表（或数组）统计字符串中每个字符出现的次数。
+
+第二遍，只要遍历到一个只出现一次的字符，那么就返回该字符，否则在遍历结束后返回空格。
 
 <!-- tabs:start -->
 
 ### **Python3**
+
 ```python
 import collections
 
 class Solution:
     def firstUniqChar(self, s: str) -> str:
-        if s == '':
-            return ' '
-        cache = collections.OrderedDict()
+        counter = collections.Counter()
+        for c in s:
+            counter[c] += 1
         for c in s:
-            cache[c] = 1 if cache.get(c) is None else cache[c] + 1
-        for k, v in cache.items():
-            if v == 1:
-                return k
+            if counter[c] == 1:
+                return c
         return ' '
-````
+```
 
 ### **Java**
 
 ```java
 class Solution {
     public char firstUniqChar(String s) {
-        if ("".equals(s)) {
-            return ' ';
-        }
-        Map<Character, Integer> map = new LinkedHashMap<>();
-        char[] chars = s.toCharArray();
-        for (char c : chars) {
-            map.put(c, map.get(c) == null ? 1 : 1 + map.get(c));
+        int n;
+        if ((n = s.length()) == 0) return ' ';
+        int[] counter = new int[26];
+        for (int i = 0; i < n; ++i) {
+            int index = s.charAt(i) - 'a';
+            ++counter[index];
         }
-        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
-            if (entry.getValue() == 1) {
-                return entry.getKey();
-            }
+        for (int i = 0; i < n; ++i) {
+            int index = s.charAt(i) - 'a';
+            if (counter[index] == 1) return s.charAt(i);
         }
         return ' ';
     }
@@ -71,13 +73,15 @@ class Solution {
  * @return {character}
  */
 var firstUniqChar = function (s) {
-  let t = new Array(26).fill(0);
-  let code = "a".charCodeAt();
-  for (let i = 0; i < s.length; i++) {
-    t[s[i].charCodeAt() - code]++;
+  if (s.length == 0) return " ";
+  let counter = new Array(26).fill(0);
+  for (let i = 0; i < s.length; ++i) {
+    const index = s[i].charCodeAt() - "a".charCodeAt();
+    ++counter[index];
   }
-  for (let i = 0; i < s.length; i++) {
-    if (t[s[i].charCodeAt() - code] === 1) return s[i];
+  for (let i = 0; i < s.length; ++i) {
+    const index = s[i].charCodeAt() - "a".charCodeAt();
+    if (counter[index] == 1) return s[i];
   }
   return " ";
 };

lcof/面试题50. 第一个只出现一次的字符/README.md

@@ -1,17 +1,15 @@
 class Solution {
     public char firstUniqChar(String s) {
-        if ("".equals(s)) {
-            return ' ';
+        int n;
+        if ((n = s.length()) == 0) return ' ';
+        int[] counter = new int[26];
+        for (int i = 0; i < n; ++i) {
+            int index = s.charAt(i) - 'a';
+            ++counter[index];
         }
-        Map<Character, Integer> map = new LinkedHashMap<>();
-        char[] chars = s.toCharArray();
-        for (char c : chars) {
-            map.put(c, map.get(c) == null ? 1 : 1 + map.get(c));
-        }
-        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
-            if (entry.getValue() == 1) {
-                return entry.getKey();
-            }
+        for (int i = 0; i < n; ++i) {
+            int index = s.charAt(i) - 'a';
+            if (counter[index] == 1) return s.charAt(i);
         }
         return ' ';
     }

lcof/面试题50. 第一个只出现一次的字符/Solution.java

@@ -3,13 +3,15 @@
  * @return {character}
  */
 var firstUniqChar = function (s) {
-  let t = new Array(26).fill(0);
-  let code = "a".charCodeAt();
-  for (let i = 0; i < s.length; i++) {
-    t[s[i].charCodeAt() - code]++;
+  if (s.length == 0) return " ";
+  let counter = new Array(26).fill(0);
+  for (let i = 0; i < s.length; ++i) {
+    const index = s[i].charCodeAt() - "a".charCodeAt();
+    ++counter[index];
   }
-  for (let i = 0; i < s.length; i++) {
-    if (t[s[i].charCodeAt() - code] === 1) return s[i];
+  for (let i = 0; i < s.length; ++i) {
+    const index = s[i].charCodeAt() - "a".charCodeAt();
+    if (counter[index] == 1) return s[i];
   }
   return " ";
-};
+};

lcof/面试题50. 第一个只出现一次的字符/Solution.js

@@ -2,12 +2,10 @@
 
 class Solution:
     def firstUniqChar(self, s: str) -> str:
-        if s == '':
-            return ' '
-        cache = collections.OrderedDict()
+        counter = collections.Counter()
         for c in s:
-            cache[c] = 1 if cache.get(c) is None else cache[c] + 1
-        for k, v in cache.items():
-            if v == 1:
-                return k
+            counter[c] += 1
+        for c in s:
+            if counter[c] == 1:
+                return c
         return ' '

lcof/面试题50. 第一个只出现一次的字符/Solution.py

@@ -34,7 +34,7 @@
 class Solution:
     def constructArr(self, a: List[int]) -> List[int]:
         n = len(a)
-        output = [1 for _ in a]
+        output = [1] * n
         left = right = 1
         for i in range(n):
             output[i] = left

lcof/面试题66. 构建乘积数组/README.md

@@ -1,12 +1,12 @@
 class Solution:
     def constructArr(self, a: List[int]) -> List[int]:
         n = len(a)
-        output = [1 for _ in a]
+        output = [1] * n
         left = right = 1
         for i in range(n):
             output[i] = left
             left *= a[i]
         for i in range(n - 1, -1, -1):
             output[i] *= right
             right *= a[i]
-        return output
+        return output

lcof/面试题66. 构建乘积数组/Solution.py

@@ -72,7 +72,20 @@ Result 表：
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    customer_id
+FROM
+    Customer
+GROUP BY
+    customer_id
+HAVING
+    COUNT(DISTINCT(product_key)) = (
+        SELECT
+            COUNT(1)
+        FROM
+            Product
+    );
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1045.Customers Who Bought All Products/README.md

@@ -77,7 +77,20 @@ The customers who bought all the products (5 and 6) are customers with id 1 and
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    customer_id
+FROM
+    Customer
+GROUP BY
+    customer_id
+HAVING
+    COUNT(DISTINCT(product_key)) = (
+        SELECT
+            COUNT(1)
+        FROM
+            Product
+    );
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1045.Customers Who Bought All Products/README_EN.md

@@ -0,0 +1,14 @@
+# Write your MySQL query statement below
+SELECT
+    customer_id
+FROM
+    Customer
+GROUP BY
+    customer_id
+HAVING
+    COUNT(DISTINCT(product_key)) = (
+        SELECT
+            COUNT(1)
+        FROM
+            Product
+    );

solution/1000-1099/1045.Customers Who Bought All Products/Solution.sql

@@ -52,12 +52,20 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+`GROUP BY` + `HAVING` 解决。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    actor_id, director_id
+FROM
+    ActorDirector
+GROUP BY actor_id, director_id
+HAVING count(1) >= 3;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README.md

@@ -90,12 +90,20 @@ The only pair is (1, 1) where they cooperated exactly 3 times.
 
 ## Solutions
 
+Use `GROUP BY` & `HAVING`.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    actor_id, director_id
+FROM
+    ActorDirector
+GROUP BY actor_id, director_id
+HAVING count(1) >= 3;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README_EN.md

@@ -0,0 +1,7 @@
+# Write your MySQL query statement below
+SELECT
+    actor_id, director_id
+FROM
+    ActorDirector
+GROUP BY actor_id, director_id
+HAVING count(1) >= 3;