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+# Algorithm Homework 1
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+- ## 作业要求
+ 3道上课讲的LeetCode题目1，69，70
+
+- ## Leetcode problem 1 (两数之和)
+ - 算法思路：
+ 用字典记录数字，以便在找到成对的另一个数的时候，可以快速的检索到前一个数的索引
+
+ - Leetcode提交结果
+ 
+ - 代码：
+ ```python
+ def twoSum(self, nums: List[int], target: int) -> List[int]:
+ myDict = {}
+ for index, num in enumerate(nums):
+ if num in myDict:
+ return [myDict[num], index]
+ else:
+ myDict[target - num] = index
+ ```
+
+- ## Leetcode problem 69 (x 的平方根)
+ - 算法思路：二分法
+
+ - Leetcode提交结果
+ 
+ - 代码：
+ ```python
+ def mySqrt(self, x: int) -> int:
+ if x in [1, 2, 3]:
+ return 1
+ left = 0
+ right = x // 2
+ while True:
+ mid = (left + right) // 2
+ if left + 1 >= right:
+ break
+ if mid < x / mid: # mid to small
+ left = mid
+ else:
+ right = mid
+ return right if right * right <= x else left
+ ```
+
+- ## Leetcode problem 70 (爬楼梯)
+ - 算法思路：动态规划。由于当前状态只取决于前两个状态，所以dp table可以精简为只用两个数表示
+
+ - Leetcode提交结果
+ 
+ - 代码：
+ ```python
+ def climbStairs(self, n: int) -> int:
+ pre, cur = 0, 1
+ for i in range(1, n + 1):
+ pre, cur = cur, pre + cur
+ return cur
+ ```
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+# Algorithm Homework 2
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+
+
+## 题解
+
+- ### Leetcode problem 7（[整数反转](https://leetcode-cn.com/problems/reverse-integer/)）
+
+ - **算法思路：**
+
+ - 首先记录是否为负数，接着将输入值取整
+ - 用模10取余的方式依次得到反转的每一位，然后拼接到一个 *result* 当中
+ - 最后判断是否溢出，溢出则返回0，不溢出则返回 *result* 或 *-result*
+
+ - **Leetcode提交结果：**
+
+ 
+
+ - **代码：**
+
+ ```python
+ INT_MAX_VAL = 2147483647
+ INT_MIN_VAL = -2147483648
+ class Solution:
+ def reverse(self, x: int) -> int:
+ isNegitive = -1 if x < 0 else 1
+ result, x = 0, abs(x)
+ while x > 0:
+ result = result * 10 + x % 10
+ x //= 10
+ return 0 if result > INT_MAX_VAL or result < INT_MIN_VAL else isNegitive * result
+ ```
+
+- ### Leetcode problem 13（[罗马数字转整数](https://leetcode-cn.com/problems/roman-to-integer/)）
+
+ - **算法思路：**
+
+ - 首先对于 *M、D、L、V*，可直接返回对应代表的数值
+ - 对于 *C、X、I*，则如果右边出现可与之搭配成9 × 10n 或者 4 × 10n（n = 0, 1, 2）的情况，返回负的对应数值，否则返回正的对应数值
+
+ - **Leetcode提交结果：**
+
+ 
+
+ - **代码：**
+
+ ```python
+ class Solution:
+ def romanToInt(self, s: str) -> int:
+ result, length = 0, len(s)
+ for i in range(0, length):
+ if s[i] == 'M':
+ result += 1000
+ elif s[i] == 'D':
+ result += 500
+ elif s[i] == 'C':
+ if i < length - 1 and (s[i + 1] == 'M' or s[i + 1] == 'D'):
+ result -= 100
+ else:
+ result += 100
+ elif s[i] == 'L':
+ result += 50
+ elif s[i] == 'X':
+ if i < length - 1 and (s[i + 1] == 'L' or s[i + 1] == 'C'):
+ result -= 10
+ else:
+ result += 10
+ elif s[i] == 'V':
+ result += 5
+ elif s[i] == 'I':
+ if i < length - 1 and (s[i + 1] == 'V' or s[i + 1] == 'X'):
+ result -= 1
+ else:
+ result += 1
+ return result
+ ```
+
+- ### Leetcode problem 3（[加一](https://leetcode-cn.com/problems/plus-one/)）
+
+ - **算法思路：**
+
+ - 从后往前遍历，每次 + 1，如果和等于10，则发生进位，否则算法返回
+ - 遍历结束发现 *tmp* 仍然不为0，则表示数组的值为全9，需要在数组前添加一个元素
+
+ - **Leetcode提交结果：**
+
+ 
+
+ - **代码：**
+
+ ```python
+ class Solution:
+ def plusOne(self, digits: List[int]) -> List[int]:
+ tmp = digits[len(digits) - 1] + 1
+ digits[len(digits) - 1], tmp, index = tmp % 10, tmp // 10, len(digits) - 2
+ while tmp != 0 and index >= 0:
+ tmp = digits[index] + 1
+ digits[index], tmp = tmp % 10, tmp // 10
+ index -= 1
+ if tmp != 0:
+ digits.insert(0, 1)
+ return digits
+ ```
+
+
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+# Algorithm Homework 3
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+Leetcode: 35,58,83
+
+## 题解：
+
+- ### Leetcode problem 35（[搜索插入位置](https://leetcode-cn.com/problems/search-insert-position/)）
+
+ - **算法思路：**
+
+ - 尝试使用二分法找到target的索引，找到则返回
+ - 没有找到则判断target与左右索引的大小关系，返回应该插入的位置
+
+ - **Leetcode提交结果：**
+
+ 
+
+ - **代码：**
+
+ ```python
+ class Solution:
+ def searchInsert(self, nums: List[int], target: int) -> int:
+ left, right = 0, len(nums) - 1
+ while left < right - 1:
+ pos = (left + right) // 2
+ if nums[pos] > target:
+ right = pos
+ elif nums[pos] < target:
+ left = pos
+ else:
+ return pos
+ if nums[left] == target or target < nums[left]:
+ return left
+ elif nums[right] == target or target < nums[right]:
+ return right
+ else:
+ return right + 1
+ ```
+
+- ### Leetcode problem 58（[最后一个单词的长度](https://leetcode-cn.com/problems/length-of-last-word/)）
+
+ - **算法思路：**
+
+ - 思路1：使用`split`将字符串用空格分割成数组，然后反序遍历，取第一个非空字符串的长度，无则返回0；
+ - 思路2：使用`strip`截掉字符串前后的空格，然后反序遍历统计最后一个单次的长度，无则返回0。
+
+ - **Leetcode提交结果：**
+
+ 
+
+ - **代码：**
+
+ ```python
+ # 思路1代码
+ class Solution:
+ def lengthOfLastWord(self, s: str) -> int:
+ strs = s.split(' ')
+ for i in range(0, len(strs))[::-1]:
+ if len(strs[i]) > 0: return len(strs[i])
+ return 0
+
+ # 思路2代码（性能更优）
+ class Solution:
+ def lengthOfLastWord(self, s: str) -> int:
+ s = s.strip()
+ count = 0
+ for i in range(0, len(s))[::-1]:
+ if s[i] == ' ':
+ return count
+ else:
+ count += 1
+ return count
+ ```
+
+- ### Leetcode problem 83（[删除排序链表中的重复元素](https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/)）
+
+ - **算法思路：**
+
+ - 用一个set记录当前遍历过的值
+ - 遍历到每一个节点，用set检查当前值是否存在于set，如果存在则删除该节点
+
+ - **Leetcode提交结果：**
+
+ 
+
+ - **代码：**
+
+ ```python
+ class Solution:
+ def deleteDuplicates(self, head: ListNode) -> ListNode:
+ if head == None:
+ return head
+ s, pre, cur = set(), head, head.next
+ s.add(pre.val)
+ while cur is not None:
+ if cur.val in s:
+ pre.next, cur = cur.next, cur.next
+ continue
+ else:
+ s.add(cur.val)
+ pre, cur = cur, cur.next
+ return head
+ ```
+
+
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# algorithm-homework
+
+- [第一次作业](AlgorithmHomework1.md)
+- [第二次作业](AlgorithmHomework2.md)
+- [第三次作业](AlgorithmHomework3.md)
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+ 算法课作业-阙建明
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