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111 lines
3.4 KiB
Markdown
111 lines
3.4 KiB
Markdown
# Algorithm Homework 3
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> 姓名：阙建明
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>
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> 学号：1901213139
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## 作业要求：
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Leetcode: 35,58,83
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## 题解：
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- ### Leetcode problem 35（[搜索插入位置](https://leetcode-cn.com/problems/search-insert-position/)）
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- **算法思路：**
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- 尝试使用二分法找到target的索引，找到则返回
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- 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
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- **Leetcode提交结果：**
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- **代码：**
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```python
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class Solution:
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def searchInsert(self, nums: List[int], target: int) -> int:
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left, right = 0, len(nums) - 1
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while left < right - 1:
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pos = (left + right) // 2
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if nums[pos] > target:
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right = pos
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elif nums[pos] < target:
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left = pos
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else:
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return pos
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if nums[left] == target or target < nums[left]:
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return left
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elif nums[right] == target or target < nums[right]:
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return right
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else:
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return right + 1
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```
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- ### Leetcode problem 58（[最后一个单词的长度](https://leetcode-cn.com/problems/length-of-last-word/)）
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- **算法思路：**
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- 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；
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- 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。
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- **Leetcode提交结果：**
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- **代码：**
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```python
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# 思路1代码
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class Solution:
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def lengthOfLastWord(self, s: str) -> int:
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strs = s.split(' ')
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for i in range(0, len(strs))[::-1]:
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if len(strs[i]) > 0: return len(strs[i])
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return 0
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# 思路2代码（性能更优）
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class Solution:
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def lengthOfLastWord(self, s: str) -> int:
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s = s.strip()
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count = 0
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for i in range(0, len(s))[::-1]:
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if s[i] == ' ':
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return count
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else:
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count += 1
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return count
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```
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- ### 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/)）
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- **算法思路：**
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- 用一个set记录当前遍历过的值
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- 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
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- **Leetcode提交结果：**
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- **代码：**
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```python
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class Solution:
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def deleteDuplicates(self, head: ListNode) -> ListNode:
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if head == None:
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return head
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s, pre, cur = set(), head, head.next
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s.add(pre.val)
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while cur is not None:
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if cur.val in s:
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pre.next, cur = cur.next, cur.next
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continue
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else:
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s.add(cur.val)
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pre, cur = cur, cur.next
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return head
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```
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