feat: add solutions to lcci problems: 01.01~01.06

lcci/01.01.Is Unique/README.md

@@ -35,7 +35,7 @@
 
 因此，我们可以使用一个 $32$ 位整数 `mask` 的每一位来表示字符串中的每一个字符是否出现过。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串长度。
+时间复杂度 $O(n)$，其中 $n$ 为字符串长度。空间复杂度 $O(1)$。
 
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lcci/01.01.Is Unique/README_EN.md

@@ -34,6 +34,14 @@
 
 ## Solutions
 
+**Solution 1: Bit Manipulation**
+
+Based on the examples, we can assume that the string only contains lowercase letters (which is confirmed by actual verification).
+
+Therefore, we can use each bit of a $32$-bit integer `mask` to represent whether each character in the string has appeared.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
+
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 ### **Python3**

lcci/01.02.Check Permutation/README.md

@@ -32,7 +32,7 @@
 
 **方法一：数组或哈希表**
 
-先判断两个字符串的长度是否相等，若不相等则直接返回 `false`。
+我们先判断两个字符串的长度是否相等，若不相等则直接返回 `false`。
 
 然后用一个数组或哈希表统计字符串 $s1$ 中字符出现的次数。
 
@@ -46,9 +46,9 @@
 
 **方法二：排序**
 
-按照字典序对两个字符串进行排序，然后比较两个字符串是否相等。
+我们也按照字典序对两个字符串进行排序，然后比较两个字符串是否相等。
 
-时间复杂度 $O(n\log n)$，空间复杂度 $O(1)$。
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。
 
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lcci/01.02.Check Permutation/README_EN.md

@@ -34,6 +34,26 @@
 
 ## Solutions
 
+**Solution 1: Array or Hash Table**
+
+First, we check whether the lengths of the two strings are equal. If they are not equal, we directly return `false`.
+
+Then, we use an array or hash table to count the occurrence of each character in string $s1$.
+
+Next, we traverse the other string $s2$. For each character we encounter, we decrement its corresponding count. If the count after decrementing is less than $0$, it means that the occurrence of characters in the two strings is different, so we directly return `false`.
+
+Finally, after traversing string $s2$, we return `true`.
+
+Note: In this problem, all test case strings only contain lowercase letters, so we can directly create an array of length $26$ for counting.
+
+The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, $C=26$.
+
+**Solution 2: Sorting**
+
+We can also sort the two strings in lexicographical order, and then compare whether the two strings are equal.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
+
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 ### **Python3**

lcci/01.03.String to URL/README.md

@@ -33,13 +33,13 @@
 
 直接利用 `replace` 将所有 ` ` 替换为 `%20`：
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串长度。
 
 **方法二：模拟**
 
 遍历字符串每个字符 $c$，遇到空格则将 `%20` 添加到结果中，否则添加 $c$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串长度。
 
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lcci/01.03.String to URL/README_EN.md

@@ -40,6 +40,18 @@ The missing numbers are [5,6,8,...], hence the third missing number is 8.
 
 ## Solutions
 
+**Solution 1: Using `replace()` function**
+
+Directly use `replace` to replace all ` ` with `%20`:
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
+
+**Solution 2: Simulation**
+
+Traverse each character $c$ in the string. When encountering a space, add `%20` to the result, otherwise add $c$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
+
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 ### **Python3**

lcci/01.04.Palindrome Permutation/README.md

@@ -27,7 +27,7 @@
 
 **方法一：哈希表**
 
-我们用哈希表 `cnt` 存储每个字符出现的次数。若次数为奇数的字符超过 $1$ 个，则不是回文排列。
+我们用哈希表 $cnt$ 存储每个字符出现的次数。若次数为奇数的字符超过 $1$ 个，则不是回文排列。
 
 时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串长度。
 

lcci/01.04.Palindrome Permutation/README_EN.md

@@ -20,6 +20,12 @@
 
 ## Solutions
 
+**Solution 1: Hash Table**
+
+We use a hash table $cnt$ to store the occurrence count of each character. If more than $1$ character has an odd count, then it is not a palindrome permutation.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
+
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 ### **Python3**