feat: update solutions to lc problems: No.2710~2712 (#3120)

Author: yanglbme

solution/2600-2699/2664.The Knight’s Tour/README_EN.md

@@ -59,7 +59,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Backtracking
+
+We create a two-dimensional array $g$, used to record the knight's movement order, initially $g[r][c] = -1$, and all other positions are set to $-1$ as well. Additionally, we need a variable $ok$ to record whether a solution has been found.
+
+Next, we start depth-first search from $(r, c)$. Each time we search position $(i, j)$, we first check if $g[i][j]$ equals $m \times n - 1$. If so, it means we have found a solution, then we set $ok$ to `true` and return. Otherwise, we enumerate the knight's eight possible movement directions to position $(x, y)$. If $0 \leq x < m$, $0 \leq y < n$, and $g[x][y]=-1$, then we update $g[x][y]$ to $g[i][j]+1$, and recursively search position $(x, y)$. If after the search, the variable $ok$ is `true`, we return directly. Otherwise, we reset $g[x][y]$ to $-1$ and continue searching in other directions.
+
+Finally, return the two-dimensional array $g$.
+
+The time complexity is $O(8^{m \times n})$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the integers given in the problem.
 
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solution/2700-2799/2710.Remove Trailing Zeros From a String/README.md

@@ -140,30 +140,4 @@ impl Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn remove_trailing_zeros(num: String) -> String {
-        num.chars()
-            .rev()
-            .skip_while(|&c| c == '0')
-            .collect::<String>()
-            .chars()
-            .rev()
-            .collect::<String>()
-    }
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2700-2799/2710.Remove Trailing Zeros From a String/README_EN.md

@@ -52,7 +52,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Traversal
+
+We can traverse the string from the end to the beginning, stopping when we encounter the first character that is not `0`. Then, we return the substring from the beginning to this character.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. Ignoring the space consumed by the answer string, the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -136,30 +140,4 @@ impl Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Rust
-
-```rust
-impl Solution {
-    pub fn remove_trailing_zeros(num: String) -> String {
-        num.chars()
-            .rev()
-            .skip_while(|&c| c == '0')
-            .collect::<String>()
-            .chars()
-            .rev()
-            .collect::<String>()
-    }
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2700-2799/2710.Remove Trailing Zeros From a String/Solution2.rs

@@ -78,7 +78,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：模拟
+
+我们可以按照题目描述的流程模拟，计算出每个单元格的左上角对角线上不同值的数量 $tl$ 和右下角对角线上不同值的数量 $br$，然后计算它们的差值 $|tl - br|$。
+
+时间复杂度 $O(m \times n \times \min(m, n))$，空间复杂度 $O(m \times n)$。
 
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solution/2700-2799/2711.Difference of Number of Distinct Values on Diagonals/README.md

@@ -170,7 +170,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We can simulate the process described in the problem statement, calculating the number of distinct values on the top-left diagonal $tl$ and the bottom-right diagonal $br$ for each cell, then compute their difference $|tl - br|$.
+
+The time complexity is $O(m \times n \times \min(m, n))$, and the space complexity is $O(m \times n)$.
 
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solution/2700-2799/2711.Difference of Number of Distinct Values on Diagonals/README_EN.md

@@ -68,7 +68,15 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：贪心
+
+根据题目描述，如果 $s[i] \neq s[i - 1]$，那么一定要执行操作，否则无法使所有字符相等。
+
+我们要么选择将 $s[0..i-1]$ 的字符全部反转，反转的成本为 $i$；要么选择将 $s[i..n-1]$ 的字符全部反转，反转的成本为 $n - i$。取两者中的最小值即可。
+
+我们遍历字符串 $s$，将所有需要反转的字符的成本相加，即可得到最小成本。
+
+时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
 
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solution/2700-2799/2712.Minimum Cost to Make All Characters Equal/README.md

@@ -67,7 +67,15 @@ The total cost to make all characters equal is 9. It can be shown that 9 is the
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy Algorithm
+
+According to the problem description, if $s[i] \neq s[i - 1]$, an operation must be performed; otherwise, it's impossible to make all characters equal.
+
+We can either choose to reverse all characters from $s[0..i-1]$, with a cost of $i$, or reverse all characters from $s[i..n-1]$, with a cost of $n - i$. We take the minimum of the two.
+
+By iterating through the string $s$ and summing up the costs of all characters that need to be reversed, we can obtain the minimum cost.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
 
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