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/python/chapter_dynamic_programming/climbing_stairs_backtrack.py

@@ -22,7 +22,7 @@ def backtrack(choices: list[int], state: int, n: int, res: list[int]) -> int:
 
 def climbing_stairs_backtrack(n: int) -> int:
     """爬楼梯：回溯"""
-    choices = [1, 2]  # 可选择向上爬 1 或 2 阶
+    choices = [1, 2]  # 可选择向上爬 1 阶或 2 阶
     state = 0  # 从第 0 阶开始爬
     res = [0]  # 使用 res[0] 记录方案数量
     backtrack(choices, state, n, res)

codes/python/chapter_dynamic_programming/coin_change.py

@@ -14,7 +14,7 @@ def coin_change_dp(coins: list[int], amt: int) -> int:
     # 状态转移：首行首列
     for a in range(1, amt + 1):
         dp[0][a] = MAX
-    # 状态转移：其余行列
+    # 状态转移：其余行和列
     for i in range(1, n + 1):
         for a in range(1, amt + 1):
             if coins[i - 1] > a:

codes/python/chapter_dynamic_programming/edit_distance.py

@@ -62,7 +62,7 @@ def edit_distance_dp(s: str, t: str) -> int:
         dp[i][0] = i
     for j in range(1, m + 1):
         dp[0][j] = j
-    # 状态转移：其余行列
+    # 状态转移：其余行和列
     for i in range(1, n + 1):
         for j in range(1, m + 1):
             if s[i - 1] == t[j - 1]:

codes/python/chapter_dynamic_programming/knapsack.py

@@ -7,10 +7,10 @@ Author: Krahets (krahets@163.com)
 
 def knapsack_dfs(wgt: list[int], val: list[int], i: int, c: int) -> int:
     """0-1 背包：暴力搜索"""
-    # 若已选完所有物品或背包无容量，则返回价值 0
+    # 若已选完所有物品或背包无剩余容量，则返回价值 0
     if i == 0 or c == 0:
         return 0
-    # 若超过背包容量，则只能不放入背包
+    # 若超过背包容量，则只能选择不放入背包
     if wgt[i - 1] > c:
         return knapsack_dfs(wgt, val, i - 1, c)
     # 计算不放入和放入物品 i 的最大价值
@@ -24,13 +24,13 @@ def knapsack_dfs_mem(
     wgt: list[int], val: list[int], mem: list[list[int]], i: int, c: int
 ) -> int:
     """0-1 背包：记忆化搜索"""
-    # 若已选完所有物品或背包无容量，则返回价值 0
+    # 若已选完所有物品或背包无剩余容量，则返回价值 0
     if i == 0 or c == 0:
         return 0
     # 若已有记录，则直接返回
     if mem[i][c] != -1:
         return mem[i][c]
-    # 若超过背包容量，则只能不放入背包
+    # 若超过背包容量，则只能选择不放入背包
     if wgt[i - 1] > c:
         return knapsack_dfs_mem(wgt, val, mem, i - 1, c)
     # 计算不放入和放入物品 i 的最大价值

codes/python/chapter_dynamic_programming/min_path_sum.py

@@ -55,7 +55,7 @@ def min_path_sum_dp(grid: list[list[int]]) -> int:
     # 状态转移：首列
     for i in range(1, n):
         dp[i][0] = dp[i - 1][0] + grid[i][0]
-    # 状态转移：其余行列
+    # 状态转移：其余行和列
     for i in range(1, n):
         for j in range(1, m):
             dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j]