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Commit f9bf911

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feat: update solutions to lcof questions
更新剑指 Offer 题解描述
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‎lcof/面试题58 - I. 翻转单词顺序/README.md‎

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**注意:** 此题对比原题有改动
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## 解法
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按空格分割字符串后逆序。
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### Python3
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```python
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class Solution:

‎lcof/面试题60. n个骰子的点数/README.md‎

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假设 `dp[i][j]` 表示扔 i 个骰子,出现点数之和 j 的次数。n 个骰子,所以 i 的范围在 `1~n`,j 的范围在 `[1, 6n]`。
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单看第 n 枚骰子,它的点数可能为 `1,2,3,...,6`,因此扔完 n 枚骰子后点数之和 j 出现的次数,可以由扔完 n-1 枚骰子后,对应点数 j−1,j−2,j−3,...,j−6 出现的次数之和转化过来。即:
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单看第 n 枚骰子,它的点数可能为 `1,2,3,...,6`,因此扔完 n 枚骰子后点数之和 j 出现的次数,可以由扔完 n-1 枚骰子后,对应点数 `j−1,j−2,j−3,...,j−6` 出现的次数之和转化过来。即:
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```
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for (第n枚骰子的点数 i = 1; i <= 6; i ++) {
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dp[n][j] += dp[n-1][j - i]
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}
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```
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扔 1 枚骰子,点数可能是 1,2,3,4,5,6,且每个点数出现的次数均为 1。所以初始化如下:
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扔 1 枚骰子,点数可能是 `1,2,3,4,5,6`,且每个点数出现的次数均为 1。所以初始化如下:
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```
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for (int j = 1; j <= 6; ++j) {

‎lcof/面试题64. 求1+2+...+n/README.md‎

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- `1 <= n <= 10000`
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## 解法
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递归,结合**逻辑与**短路运算符求解。
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### Python3
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```python
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class Solution:

‎lcof/面试题65. 不用加减乘除做加法/README.md‎

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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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对两数进行 `^` 异或运算,得到不进位的和;对两数进行 `&` 与运算,得到进位。循环,直至进位为 0。
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- 对两数进行按位 `^` 异或运算,得到不进位的和;
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- 对两数进行按位 `&` 与运算,然后左移一位,得到进位;
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- 循环,直至进位为 0。
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### Python3
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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由于 python int 是无限长整型,左移不会自动溢出,因此需要特殊处理。
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由于 python `int` 是无限长整型,左移不会自动溢出,因此需要特殊处理。
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```python
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class Solution:

‎lcof/面试题68 - II. 二叉树的最近公共祖先/README.md‎

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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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if left is None:
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return right
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if right is None:
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return left
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return root
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return right if left is None else (left if right is None else root)
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```
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### Java

‎lcof/面试题68 - II. 二叉树的最近公共祖先/Solution.py‎

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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:
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if root is None or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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if left is None:
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return right
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if right is None:
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return left
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return root
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return right if left is None else (left if right is None else root)

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