Merge 3e40c1153d into 894e3d536b

打印的小错误

codes/cpp/chapter_tree/array_binary_tree.cpp

@@ -115,9 +115,9 @@ int main() {
     int i = 1;
     int l = abt.left(i), r = abt.right(i), p = abt.parent(i);
     cout << "\n当前节点的索引为 " << i << "，值为 " << abt.val(i) << "\n";
-    cout << "其左子节点的索引为 " << l << "，值为 " << (l != INT_MAX ? to_string(abt.val(l)) : "nullptr") << "\n";
+    cout << "其左子节点的索引为 " << l << "，值为 " << (abt.val(l) != INT_MAX ? to_string(abt.val(l)) : "nullptr") << "\n";
-    cout << "其右子节点的索引为 " << r << "，值为 " << (r != INT_MAX ? to_string(abt.val(r)) : "nullptr") << "\n";
+    cout << "其右子节点的索引为 " << r << "，值为 " << (abt.val(r) != INT_MAX ? to_string(abt.val(r)) : "nullptr") << "\n";
-    cout << "其父节点的索引为 " << p << "，值为 " << (p != INT_MAX ? to_string(abt.val(p)) : "nullptr") << "\n";
+    cout << "其父节点的索引为 " << p << "，值为 " << (abt.val(p) != INT_MAX ? to_string(abt.val(p)) : "nullptr") << "\n";
 
     // 遍历树
     vector<int> res = abt.levelOrder();

en/docs/chapter_searching/replace_linear_by_hashing.md

@@ -1,6 +1,6 @@
 # Hash optimization strategies
 
-In algorithm problems, **we often reduce the time complexity of algorithms by replacing linear search with hash search**. Let's use an algorithm problem to deepen understanding.
+In algorithm problems, **we often reduce an algorithm's time complexity by replacing a linear search with a hash-based search**. Let's use an algorithm problem to deepen the understanding.
 
 !!! question
 
@@ -8,7 +8,7 @@ In algorithm problems, **we often reduce the time complexity of algorithms by re
 
 ## Linear search: trading time for space
 
-Consider traversing all possible combinations directly. As shown in the figure below, we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
+Consider traversing through all possible combinations directly. As shown in the figure below, we initiate a nested loop, and in each iteration, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
 
 ![Linear search solution for two-sum problem](replace_linear_by_hashing.assets/two_sum_brute_force.png)
 
@@ -18,11 +18,11 @@ The code is shown below:
 [file]{two_sum}-[class]{}-[func]{two_sum_brute_force}
 ```
 
-This method has a time complexity of $O(n^2)$ and a space complexity of $O(1)$, which is very time-consuming with large data volumes.
+This method has a time complexity of $O(n^2)$ and a space complexity of $O(1)$, which becomes very time-consuming with large data volumes.
 
 ## Hash search: trading space for time
 
-Consider using a hash table, with key-value pairs being the array elements and their indices, respectively. Loop through the array, performing the steps shown in the figure below each round.
+Consider using a hash table, where the key-value pairs are the array elements and their indices, respectively. Loop through the array, performing the steps shown in the figure below during each iteration.
 
 1. Check if the number `target - nums[i]` is in the hash table. If so, directly return the indices of these two elements.
 2. Add the key-value pair `nums[i]` and index `i` to the hash table.
@@ -42,6 +42,6 @@ The implementation code is shown below, requiring only a single loop:
 [file]{two_sum}-[class]{}-[func]{two_sum_hash_table}
 ```
 
-This method reduces the time complexity from $O(n^2)$ to $O(n)$ by using hash search, greatly improving the running efficiency.
+This method reduces the time complexity from $O(n^2)$ to $O(n)$ by using hash search, significantly improving the running efficiency.
 
 As it requires maintaining an additional hash table, the space complexity is $O(n)$. **Nevertheless, this method has a more balanced time-space efficiency overall, making it the optimal solution for this problem**.