增加9习题

data/3.dailycode高阶/3.python/1.exercises/solution.md

@@ -21,6 +21,42 @@
   
 </p>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+def maze(m, n, route, pos, export):
+	"""走迷宫
+	m	   - 迷宫数组，列表
+	n	   - 迷宫阶数
+	route   - 可能的路线，列表
+	pos	 - 当前位置，元组
+	export  - 出口位置，元组
+	"""
+	route.append(pos)
+	if pos == export:
+		print(route)
+	if pos[0] > 0 and m[pos[0]-1][pos[1]] == 0 and (pos[0]-1,pos[1]) not in route:
+		maze(m, n, route[:], (pos[0]-1,pos[1]), export)
+	if pos[0] < n-1 and m[pos[0]+1][pos[1]] == 0 and (pos[0]+1,pos[1]) not in route:
+		maze(m, n, route[:], (pos[0]+1,pos[1]), export)
+	if pos[1] > 0 and m[pos[0]][pos[1]-1] == 0 and (pos[0],pos[1]-1) not in route:
+		maze(m, n, route[:], (pos[0],pos[1]-1), export)
+	________________________________
+m = [
+	[1,1,1,0,1,1,1,1,1,1], 
+	[1,0,0,0,0,0,0,0,1,1], 
+	[1,0,1,1,1,1,1,0,0,1], 
+	[1,0,1,0,0,0,0,1,0,1], 
+	[1,0,1,0,1,1,0,0,0,1], 
+	[1,0,0,1,1,0,1,0,1,1], 
+	[1,1,1,1,0,0,0,0,1,1], 
+	[1,0,0,0,0,1,1,1,0,0], 
+	[1,0,1,1,0,0,0,0,0,1], 
+	[1,1,1,1,1,1,1,1,1,1] 
+]
+maze(m, len(m), list(), (0,3), (7,9))
+```
+
 ## template
 
 ```python
@@ -61,7 +97,8 @@ maze(m, len(m), list(), (0,3), (7,9))
 ## 答案
 
 ```python
-
+if pos[1] < n-1 and m[pos[0]][pos[1]+1] == 0 and (pos[0],pos[1]+1) not in route:
+	maze(m, n, route[:], (pos[0],pos[1]+1), export)
 ```
 
 ## 选项
@@ -69,17 +106,20 @@ maze(m, len(m), list(), (0,3), (7,9))
 ### A
 
 ```python
-
+if pos[1] < n-1 and m[pos[0]][pos[1]+1] == 0 and (pos[0],pos[1]+1) not in route:
+	maze(m, n, route[:], (pos[0],pos[1]), export)
 ```
 
 ### B
 
 ```python
-
+if pos[1] < n-1 and m[pos[0]][pos[1]+1] == 0 and (pos[0],pos[1]+1) not in route:
+	maze(m, n, route[:], (pos[0],pos[1]-1), export)
 ```
 
 ### C
 
 ```python
-
+if pos[1] < n-1 and m[pos[0]][pos[1]+1] == 0 and (pos[0],pos[1]+1) not in route:
+	maze(m, n, route[:], (pos[0],pos[0]+1), export)
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/2.exercises/solution.md

@@ -52,6 +52,40 @@
     </div>
 </div>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def solveSudoku(self, board):
+			def isvaild(i,j):
+				for m in range(9):
+					if m!=i and board[m][j]==board[i][j]:
+						return False
+				for n in range(9):
+					if n!=j and board[i][n]==board[i][j]:
+						return False
+				______________________
+				return True
+			def f(a,b,board):
+				for i in range(a,9):
+					for j in range(b,9):
+						if board[i][j]=='.':
+							for c in '123456789':
+								board[i][j]=c
+								if isvaild(i,j):
+									if f(a,b,board):return True
+									else: board[i][j] = '.'
+								else:board[i][j] = '.'
+							return False
+				return True
+			f(0,0,board)
+			return board
+# %%
+s = Solution()
+board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
+print(s.solveSudoku(board))
+```
+
 ## template
 
 ```python
@@ -92,7 +126,10 @@ print(s.solveSudoku(board))
 ## 答案
 
 ```python
-
+for m in range(i//3*3,i//3*3+3):
+    for n in range(j//3*3,j//3*3+3):
+        if m!=i and n!=j and board[m][n]==board[i][j]:
+            return False
 ```
 
 ## 选项
@@ -100,17 +137,26 @@ print(s.solveSudoku(board))
 ### A
 
 ```python
-
+for m in range(i//3*3,i//3*3+3):
+    for n in range(j//3*3,j//3*3+3):
+        if board[m][n]==board[i][j]:
+            return False
 ```
 
 ### B
 
 ```python
-
+for m in range(i//3*3,i//3*3+3):
+    for n in range(j//3*3,j//3*3+3):
+        if m==i and n==j and board[m][n]==board[i][j]:
+            return False
 ```
 
 ### C
 
 ```python
-
+for m in range(i//3*3,i//3*3+3):
+    for n in range(j//3*3,j//3*3+3):
+        if board[m][n]==board[i][j]:
+            return True
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/3.exercises/solution.md

@@ -2,6 +2,50 @@
 
 <p>给定 <em>n</em> 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。</p><p> </p><p><strong>示例 1：</strong></p><p><img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0042.Trapping%20Rain%20Water/images/rainwatertrap.png" style="height: 161px; width: 412px;" /></p><pre><strong>输入：</strong>height = [0,1,0,2,1,0,1,3,2,1,2,1]<strong><br />输出：</strong>6<strong><br />解释：</strong>上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。 </pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>height = [4,2,0,3,2,5]<strong><br />输出：</strong>9</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>n == height.length</code></li>	<li><code>0 <= n <= 3 * 10<sup>4</sup></code></li>	<li><code>0 <= height[i] <= 10<sup>5</sup></code></li></ul>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def trap(self, height):
+		ls = len(height)
+		if ls == 0:
+			return 0
+		res, left = 0, 0
+		while left < ls and height[left] == 0:
+			left += 1
+		pos = left + 1
+		while pos < ls:
+			if height[pos] >= height[left]:
+				res += self.rain_water(height, left, pos)
+				left = pos
+				pos += 1
+			elif pos == ls - 1:
+				max_value, max_index = 0, pos
+				for index in range(left + 1, ls):
+					if height[index] > max_value:
+						max_value = height[index]
+						max_index = index
+				res += self.rain_water(height, left, max_index)
+				left = max_index
+				pos = left + 1
+			else:
+				pos += 1
+		return res
+	def rain_water(self, height, start, end):
+		if end - start <= 1:
+			return 0
+		min_m = min(height[start], height[end])
+		_________________________
+		step = 0
+		for index in range(start + 1, end):
+			if height[index] > 0:
+				step += height[index]
+		return res - step
+if __name__ == '__main__':
+	s = Solution()
+	print (s.trap([2,6,3,8,2,7,2,5,0]))
+```
+
 ## template
 
 ```python
@@ -49,7 +93,7 @@ if __name__ == '__main__':
 ## 答案
 
 ```python
-
+res = min_m * (end - start - 1)
 ```
 
 ## 选项
@@ -57,17 +101,17 @@ if __name__ == '__main__':
 ### A
 
 ```python
-
+res = min_m * (end - start) / 2
 ```
 
 ### B
 
 ```python
-
+res = min_m * (end + start) / 2
 ```
 
 ### C
 
 ```python
-
+res = min_m * (end - start)
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/4.exercises/solution.md

@@ -2,6 +2,35 @@
 
 <p>给定 <em>n</em> 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。</p><p>求在该柱状图中，能够勾勒出来的矩形的最大面积。</p><p>&nbsp;</p><p><img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0084.Largest%20Rectangle%20in%20Histogram/images/histogram.png"></p><p><small>以上是柱状图的示例，其中每个柱子的宽度为 1，给定的高度为&nbsp;<code>[2,1,5,6,2,3]</code>。</small></p><p>&nbsp;</p><p><img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0084.Largest%20Rectangle%20in%20Histogram/images/histogram_area.png"></p><p><small>图中阴影部分为所能勾勒出的最大矩形面积，其面积为&nbsp;<code>10</code>&nbsp;个单位。</small></p><p>&nbsp;</p><p><strong>示例:</strong></p><pre><strong>输入:</strong> [2,1,5,6,2,3]<strong><br />输出:</strong> 10</pre>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def largestRectangleArea(self, heights):
+		"""
+		:type heights: List[int]
+		:rtype: int
+		"""
+		largest_rectangle = 0
+		ls = len(heights)
+		stack = [-1]
+		top, pos = 0, 0
+		for pos in range(ls):
+			while top > 0 and heights[stack[top]] > heights[pos]:
+				_____________________________
+				top -= 1
+				stack.pop()
+			stack.append(pos)
+			top += 1
+		while top > 0:
+			largest_rectangle = max(largest_rectangle, heights[stack[top]] * (ls - stack[top - 1] - 1))
+			top -= 1
+		return largest_rectangle
+if __name__ == "__main__":
+	s = Solution()
+	print (s.largestRectangleArea([2,1,5,6,2,3]))
+```
+
 ## template
 
 ```python
@@ -34,7 +63,7 @@ if __name__ == "__main__":
 ## 答案
 
 ```python
-
+largest_rectangle = max(largest_rectangle, heights[stack[top]] * (pos - stack[top - 1] - 1))
 ```
 
 ## 选项
@@ -42,17 +71,18 @@ if __name__ == "__main__":
 ### A
 
 ```python
-
+largest_rectangle = max(largest_rectangle, heights[stack[top]] * (pos - stack[top] - 1))
 ```
 
 ### B
 
 ```python
+largest_rectangle = max(largest_rectangle, heights[stack[top]] * (pos - stack[top + 1] - 1))
 
 ```
 
 ### C
 
 ```python
-
+largest_rectangle = max(largest_rectangle, heights[stack[top]] * (pos - stack[top - 1] + 1))
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/5.exercises/solution.md

@@ -2,6 +2,27 @@
 
 <p>给你一个未排序的整数数组 <code>nums</code> ，请你找出其中没有出现的最小的正整数。</p><p> </p><p><strong>进阶：</strong>你可以实现时间复杂度为 <code>O(n)</code> 并且只使用常数级别额外空间的解决方案吗？</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums = [1,2,0]<strong><br />输出：</strong>3</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums = [3,4,-1,1]<strong><br />输出：</strong>2</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>nums = [7,8,9,11,12]<strong><br />输出：</strong>1</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>0 <= nums.length <= 300</code></li>	<li><code>-2<sup>31</sup> <= nums[i] <= 2<sup>31</sup> - 1</code></li></ul>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def firstMissingPositive(self, nums):
+		ls = len(nums)
+		index = 0
+		while index < ls:
+			if nums[index] <= 0 or nums[index] > ls or nums[nums[index] - 1] == nums[index]:
+				index += 1
+			else:
+				pos = nums[index] - 1
+				nums[index], nums[pos] = nums[pos], nums[index]
+		res = 0
+		_________________________
+		return res + 1
+# %%
+s = Solution()
+print(s.firstMissingPositive(nums = [1,2,0]))
+```
+
 ## template
 
 ```python
@@ -27,7 +48,8 @@ print(s.firstMissingPositive(nums = [1,2,0]))
 ## 答案
 
 ```python
-
+while res < ls and nums[res] == res + 1:
+	res += 1
 ```
 
 ## 选项
@@ -35,17 +57,20 @@ print(s.firstMissingPositive(nums = [1,2,0]))
 ### A
 
 ```python
-
+while res < ls and nums[res] == res:
+	res += 1
 ```
 
 ### B
 
 ```python
-
+while res <= ls and nums[res] == res:
+	res += 1
 ```
 
 ### C
 
 ```python
-
+while res > ls and nums[res] == res + 1:
+	res += 1
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/6.exercises/solution.md

@@ -35,6 +35,30 @@
     </div>
 </div>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def __init__(self):
+		self.count = 0
+	def totalNQueens(self, n):
+		self.dfs(0, n, 0, 0, 0)
+		return self.count
+	def dfs(self, row, n, column, diag, antiDiag):
+		if row == n:
+			self.count += 1
+			return
+		for index in range(n):
+			isColSafe = (1 << index) & column == 0
+			isDigSafe = (1 << (n - 1 + row - index)) & diag == 0
+			isAntiDiagSafe = (1 << (row + index)) & antiDiag == 0
+			if isAntiDiagSafe and isColSafe and isDigSafe:
+				_________________________
+if __name__ == '__main__':
+	s = Solution()
+	print (s.totalNQueens(4))
+```
+
 ## template
 
 ```python
@@ -64,7 +88,9 @@ if __name__ == '__main__':
 ## 答案
 
 ```python
-
+self.dfs(row + 1,  n, (1 << index) | column,
+        (1 << (n - 1 + row - index)) | diag,
+        (1 << (row + index)) | antiDiag)
 ```
 
 ## 选项
@@ -72,17 +98,23 @@ if __name__ == '__main__':
 ### A
 
 ```python
-
+self.dfs(row + 1,  n, (index >> 1) | column,
+        (1 << (n - 1 + row - index)) | diag,
+        (1 << (row + index)) | antiDiag)
 ```
 
 ### B
 
 ```python
-
+self.dfs(row + 1,  n, (index >> 1) | column,
+        (1 << (n + 1 + row - index)) | diag,
+        (1 << (row + index)) | antiDiag)
 ```
 
 ### C
 
 ```python
-
+self.dfs(row + 1,  n, (1 << index) | column,
+        (1 << (n + 1 + row - index)) | diag,
+        (1 << (row + index)) | antiDiag)
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/7.exercises/solution.md

@@ -2,6 +2,42 @@
 
 <p>给定一个仅包含 <code>0</code> 和 <code>1</code> 、大小为 <code>rows x cols</code> 的二维二进制矩阵，找出只包含 <code>1</code> 的最大矩形，并返回其面积。</p><p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0085.Maximal%20Rectangle/images/maximal.jpg" style="width: 402px; height: 322px;" /><pre><strong>输入：</strong>matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]<strong><br />输出：</strong>6<strong><br />解释：</strong>最大矩形如上图所示。</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>matrix = []<strong><br />输出：</strong>0</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>matrix = [["0"]]<strong><br />输出：</strong>0</pre><p><strong>示例 4：</strong></p><pre><strong>输入：</strong>matrix = [["1"]]<strong><br />输出：</strong>1</pre><p><strong>示例 5：</strong></p><pre><strong>输入：</strong>matrix = [["0","0"]]<strong><br />输出：</strong>0</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>rows == matrix.length</code></li>	<li><code>cols == matrix[0].length</code></li>	<li><code>0 <= row, cols <= 200</code></li>	<li><code>matrix[i][j]</code> 为 <code>'0'</code> 或 <code>'1'</code></li></ul>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def maximalRectangle(self, matrix):
+		"""
+		:type matrix: List[List[str]]
+		:rtype: int
+		"""
+		if matrix is None or len(matrix) == 0:
+			return 0
+		ls_row, ls_col = len(matrix), len(matrix[0])
+		left, right, height = [0] * ls_col, [ls_col] * ls_col, [0] * ls_col
+		maxA = 0
+		for i in range(ls_row):
+			curr_left, curr_right = 0, ls_col
+			for j in range(ls_col):
+				if matrix[i][j] == '1':
+					height[j] += 1
+				else:
+					height[j] = 0
+			for j in range(ls_col):
+				if matrix[i][j] == '1':
+					left[j] = max(left[j], curr_left)
+				else:
+					left[j], curr_left = 0, j + 1
+			_______________________
+			for j in range(ls_col):
+				maxA = max(maxA, (right[j] - left[j]) * height[j])
+		return maxA
+# %%
+s = Solution()
+matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
+print(s.maximalRectangle(matrix))
+```
+
 ## template
 
 ```python
@@ -45,7 +81,11 @@ print(s.maximalRectangle(matrix))
 ## 答案
 
 ```python
-
+for j in range(ls_col - 1, -1, -1):
+	if matrix[i][j] == '1':
+		right[j] = min(right[j], curr_right)
+	else:
+		right[j], curr_right = ls_col, j
 ```
 
 ## 选项
@@ -53,17 +93,29 @@ print(s.maximalRectangle(matrix))
 ### A
 
 ```python
-
+for j in range(ls_col - 1, -1, -1):
+	if matrix[i][j] == '1':
+		right[j], curr_right = ls_col, j
+	else:
+		right[j] = min(right[j], curr_right)
 ```
 
 ### B
 
 ```python
-
+for j in range(ls_col - 1, -1, -1):
+	if matrix[i][j] == '1':
+		right[j], curr_right = ls_col, j
+	else:
+		right[j] = max(right[j], curr_right)
 ```
 
 ### C
 
 ```python
-
+for j in range(ls_col - 1, -1, -1):
+	if matrix[i][j] == '1':
+		right[j] = max(right[j], curr_right)
+	else:
+		right[j], curr_right = ls_col, j
 ```
\ No newline at end of file

data/3.dailycode高阶/3.python/8.exercises/solution.md

@@ -2,6 +2,30 @@
 
 <p>给你两个单词 <code>word1</code> 和 <code>word2</code>，请你计算出将 <code>word1</code> 转换成 <code>word2</code><em> </em>所使用的最少操作数 。</p><p>你可以对一个单词进行如下三种操作：</p><ul>	<li>插入一个字符</li>	<li>删除一个字符</li>	<li>替换一个字符</li></ul><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>word1 = "horse", word2 = "ros"<strong><br />输出：</strong>3<strong><br />解释：</strong>horse -> rorse (将 'h' 替换为 'r')rorse -> rose (删除 'r')rose -> ros (删除 'e')</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>word1 = "intention", word2 = "execution"<strong><br />输出：</strong>5<strong><br />解释：</strong>intention -> inention (删除 't')inention -> enention (将 'i' 替换为 'e')enention -> exention (将 'n' 替换为 'x')exention -> exection (将 'n' 替换为 'c')exection -> execution (插入 'u')</pre><p> </p><p><strong>提示：</strong></p><ul>	<li><code>0 <= word1.length, word2.length <= 500</code></li>	<li><code>word1</code> 和 <code>word2</code> 由小写英文字母组成</li></ul>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def minDistance(self, word1, word2):
+		ls_1, ls_2 = len(word1), len(word2)
+		dp = list(range(ls_1 + 1))
+		for j in range(1, ls_2 + 1):
+			pre = dp[0]
+			dp[0] = j
+			for i in range(1, ls_1 + 1):
+				temp = dp[i]
+				if word1[i - 1] == word2[j - 1]:
+					dp[i] = pre
+				else:
+					____________________________
+				pre = temp
+		return dp[ls_1]
+if __name__ == '__main__':
+	s = Solution()
+	print (s.minDistance("horse","ros"))		
+	print (s.minDistance("intention","execution"))		
+```
+
 ## template
 
 ```python
@@ -29,7 +53,7 @@ if __name__ == '__main__':
 ## 答案
 
 ```python
-
+dp[i] = min(pre + 1, dp[i] + 1, dp[i - 1] + 1)
 ```
 
 ## 选项
@@ -37,17 +61,17 @@ if __name__ == '__main__':
 ### A
 
 ```python
-
+dp[i] = min(pre + 1, dp[i] + 1, dp[i + 1] + 1)
 ```
 
 ### B
 
 ```python
-
+dp[i] = min(pre + 1, dp[i] + 1, dp[i - 1] - 1)
 ```
 
 ### C
 
 ```python
-
+dp[i] = min(pre + 1, dp[i] + 1, dp[i + 1] - 1)
 ```
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data/3.dailycode高阶/3.python/9.exercises/solution.md

@@ -64,6 +64,52 @@
     </pre>
 </div>
 
+以下程序实现了这一功能，请你填补空白处内容：
+
+```python
+class Solution(object):
+	def fullJustify(self, words, maxWidth):
+		"""
+		:type words: List[str]
+		:type maxWidth: int
+		:rtype: List[str]
+		"""
+		res = []
+		res_list = []
+		curr = []
+		count, pos = 0, 0
+		while pos < len(words):
+			word = words[pos]
+			if len(word) > maxWidth:
+				pos += 1
+			if len(word) + count + len(curr)<= maxWidth:
+				count += len(word)
+				curr.append(word)
+				pos += 1
+			else:
+				res_list.append(curr)
+				curr = []
+				count = 0
+		if len(curr) > 0:
+			res_list.append(curr)
+		for index, curr in enumerate(res_list):
+			text = ''
+			remain = sum([len(t) for t in curr])
+			if len(curr) == 1:
+				text = curr[0] + ' ' * (maxWidth - remain)
+			elif index == len(res_list) - 1:
+				text = ' '.join(curr)
+				text += ' ' * (maxWidth - remain - len(curr) + 1)
+			else:
+				___________________________
+			res.append(text)
+		return res
+if __name__ == '__main__':
+	s = Solution()
+	print (s.fullJustify(["Don't","go","around","saying","the","world","owes","you","a","living;","the","world","owes","you","nothing;","it","was","here","first."],30))
+```
+
+
 ## template
 
 ```python
@@ -119,7 +165,14 @@ if __name__ == '__main__':
 ## 答案
 
 ```python
-
+step = (maxWidth - remain) / (len(curr) - 1 )
+extra = (maxWidth - remain) % (len(curr) - 1 )
+for index in range(len(curr) - 1):
+  text += curr[index] + ' ' * int(step)
+  if extra > 0:
+    text += ' '
+    extra -= 1
+text += curr[-1]
 ```
 
 ## 选项
@@ -127,17 +180,38 @@ if __name__ == '__main__':
 ### A
 
 ```python
-
+step = (maxWidth - remain) / (len(curr) - 1 )
+extra = (maxWidth - remain) % (len(curr) - 1 )
+for index in range(len(curr) - 1):
+  text += curr[index] + ' ' * int(step)
+  if extra > 0:
+    text += ' '
+    extra += 1
+text += curr[-1]
 ```
 
 ### B
 
 ```python
-
+step = (maxWidth - remain) / (len(curr) - 1 )
+extra = (maxWidth - remain) % (len(curr) - 1 )
+for index in range(len(curr) - 1):
+  text += curr[index] + ' ' * int(step)
+  if extra > 0:
+    text += ' '
+    extra -= 1
+text += curr[1]
 ```
 
 ### C
 
 ```python
-
+step = (maxWidth - remain) / (len(curr) - 1 )
+extra = (maxWidth - remain) % (len(curr) - 1 )
+for index in range(len(curr) - 1):
+  text += curr[index] + ' ' * int(step)
+  if extra > 0:
+    text += ' '
+    extra += 1
+text += curr[1]
 ```
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