feat: Revised the book (#978)

* Sync recent changes to the revised Word.

* Revised the preface chapter

* Revised the introduction chapter

* Revised the computation complexity chapter

* Revised the chapter data structure

* Revised the chapter array and linked list

* Revised the chapter stack and queue

* Revised the chapter hashing

* Revised the chapter tree

* Revised the chapter heap

* Revised the chapter graph

* Revised the chapter searching

* Reivised the sorting chapter

* Revised the divide and conquer chapter

* Revised the chapter backtacking

* Revised the DP chapter

* Revised the greedy chapter

* Revised the appendix chapter

* Revised the preface chapter doubly

* Revised the figures

codes/rust/chapter_backtracking/n_queens.rs

@@ -21,7 +21,7 @@ fn backtrack(row: usize, n: usize, state: &mut Vec<Vec<String>>, res: &mut Vec<V
         // 计算该格子对应的主对角线和副对角线
         let diag1 = row + n - 1 - col;
         let diag2 = row + col;
-        // 剪枝：不允许该格子所在列、主对角线、副对角线存在皇后
+        // 剪枝：不允许该格子所在列、主对角线、副对角线上存在皇后
         if !cols[col] && !diags1[diag1] && !diags2[diag2] {
             // 尝试：将皇后放置在该格子
             state.get_mut(row).unwrap()[col] = "Q".into();
@@ -47,8 +47,8 @@ fn n_queens(n: usize) -> Vec<Vec<Vec<String>>> {
         state.push(row);
     }
     let mut cols = vec![false; n]; // 记录列是否有皇后
-    let mut diags1 = vec![false; 2 * n - 1]; // 记录主对角线是否有皇后
-    let mut diags2 = vec![false; 2 * n - 1]; // 记录副对角线是否有皇后
+    let mut diags1 = vec![false; 2 * n - 1]; // 记录主对角线上是否有皇后
+    let mut diags2 = vec![false; 2 * n - 1]; // 记录副对角线上是否有皇后
     let mut res: Vec<Vec<Vec<String>>> = Vec::new();
 
     backtrack(0, n, &mut state, &mut res, &mut cols, &mut diags1, &mut diags2);