docs: update solutions

Author: yanglbme

solution/0000-0099/0001.Two Sum/README_EN.md

@@ -48,7 +48,7 @@
 
 ## Solutions
 
-**Approach 1: Hash Table**
+**Solution 1: Hash Table**
 
 We can use the hash table $m$ to store the array value and the corresponding subscript.
 

solution/0000-0099/0002.Add Two Numbers/README_EN.md

@@ -42,7 +42,7 @@
 
 ## Solutions
 
-**Approach 1: Simulation**
+**Solution 1: Simulation**
 
 We traverse two linked lists $l_1$ and $l_2$ at the same time, and use the variable $carry$ to indicate whether there is a carry.
 

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README_EN.md

@@ -42,7 +42,7 @@ Notice that the answer must be a substring, "pwke" is a subsequence an
 
 ## Solutions
 
-**Approach 1: Two pointers + Hash Table**
+**Solution 1: Two pointers + Hash Table**
 
 Define a hash table to record the characters in the current window. Let $i$ and $j$ represent the start and end positions of the non-repeating substring, respectively. The length of the longest non-repeating substring is recorded by `ans`.
 

solution/0000-0099/0005.Longest Palindromic Substring/README_EN.md

@@ -32,7 +32,7 @@
 
 ## Solutions
 
-**Approach 1: Dynamic Programming**
+**Solution 1: Dynamic Programming**
 
 Set $dp[i][j]$ to indicate whether the string $s[i..j]$ is a palindrome.
 
@@ -41,7 +41,7 @@ Set $dp[i][j]$ to indicate whether the string $s[i..j]$ is a palindrome.
 
 The time complexity is $O(n^2)$ and the space complexity is $O(n^2)$. Where $n$ is the length of the string $s$.
 
-**Approach 2: Enumerate the Palindrome Center**
+**Solution 2: Enumerate the Palindrome Center**
 
 We can enumerate the palindrome center, spread to both sides, and find the longest palindrome.
 

solution/0000-0099/0006.Zigzag Conversion/README_EN.md

@@ -58,7 +58,7 @@ P     I
 
 ## Solutions
 
-**Approach 1: Simulation**
+**Solution 1: Simulation**
 
 We use a 2D array $g$ to simulate the process of the $Z$-shaped arrangement, where $g[i][j]$ represents the character in the $i$th row and the $j$th column. Initially, $i=0$, and we also define a direction variable $k$, initially $k=-1$, which means up.
 

solution/0000-0099/0007.Reverse Integer/README_EN.md

@@ -39,7 +39,7 @@
 
 ## Solutions
 
-**Approach 1: Mathematical**
+**Solution 1: Mathematical**
 
 Let $mi$ and $mx$ be $-2^{31}$ and $2^{31} - 1$ respectively. The reversed result $ans$ needs to satisfy $mi \le ans \le mx$.
 

solution/0000-0099/0012.Integer to Roman/README_EN.md

@@ -62,7 +62,7 @@ M             1000</pre>
 
 ## Solutions
 
-**Approach 1: Greedy**
+**Solution 1: Greedy**
 
 We can list all possible symbols $cs$ and corresponding values $vs$ first, then enumerate the value $vs[i]$ from large to small, and use the symbol $cs[i]$ as much as possible each time until the number $num$ becomes $0$.
 

solution/0000-0099/0013.Roman to Integer/README_EN.md

@@ -64,7 +64,7 @@ M             1000</pre>
 
 ## Solutions
 
-**Approach 1: Hash table + simulation**
+**Solution 1: Hash table + simulation**
 
 We first use a hash table $d$ to record the numerical value corresponding to each character, and then traverse the string $s$ from left to right. If the numerical value corresponding to the current character is less than the numerical value corresponding to the right character, subtract the numerical value corresponding to the current character, otherwise add the numerical value corresponding to the current character.
 

solution/0000-0099/0014.Longest Common Prefix/README_EN.md

@@ -35,7 +35,7 @@
 
 ## Solutions
 
-**Approach 1: Character Comparison**
+**Solution 1: Character Comparison**
 
 We take the first string $strs[0]$ as the benchmark, and compare the $i$th character of the string after it with the $i$th character of $strs[0]$. If it is the same, continue to compare the next character, otherwise return the first $i$ characters of $strs[0]$.
 

solution/0000-0099/0015.3Sum/README_EN.md

@@ -48,7 +48,7 @@ Notice that the order of the output and the order of the triplets does not matte
 
 ## Solutions
 
-**Approach 1: Sort + Two Pointers**
+**Solution 1: Sort + Two Pointers**
 
 We notice that the problem does not require us to return the triplet in order, so we might as well sort the array first, which makes it easy to skip duplicate elements.