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124 lines
3.8 KiB
Markdown
124 lines
3.8 KiB
Markdown
# Algorithm Homework 4
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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: 16，17，19
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## 题解：
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- ### Leetcode problem 16
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> [最接近的三数之和](https://leetcode-cn.com/problems/3sum-closest/)
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- **算法思路：**
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- 首先对nums进行从小到达排序；
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- 然后依次遍历0 ~ size - 2，固定第一个数i，对i + 1 ~ size - 1的数用双指针法处理；
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- 通过忽略已处理过的数来实现第一个数选取时的剪枝操作
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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 threeSumClosest(self, nums: List[int], target: int) -> int:
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if len(nums) < 3:
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return 0
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elif len(nums) == 3:
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return nums[0] + nums[1] + nums[2]
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nums = sorted(nums) # 先排序
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result, lastFirst = nums[0] + nums[1] + nums[2], nums[0] - 1
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for i in range(0, len(nums) - 2):
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if nums[i] == lastFirst:
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continue
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left, right, lastFirst = i + 1, len(nums) - 1, nums[i]
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while left < right:
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tmp = nums[i] + nums[left] + nums[right]
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if abs(target - tmp) < abs(target - result):
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result = tmp
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if tmp < target:
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left += 1
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elif tmp > target:
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right -= 1
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else:
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return result
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return result
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```
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- ### Leetcode problem 17
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> [电话号码的字母组合](https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/)
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- **算法思路：**
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- 采用回溯法递归遍历
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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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numMap = {'2': "abc", '3': "def", '4': "ghi", '5': "jkl", '6': "mno", '7': "pqrs", '8': "tuv", '9': "wxyz"}
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result = []
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def _letterCombinations(self, path, digits: str, index):
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if index == len(digits) - 1:
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for ch in self.numMap[digits[index]]:
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self.result.append(path + ch)
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return
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for ch in self.numMap[digits[index]]:
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path = path + ch
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self._letterCombinations(path, digits, index + 1)
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path = path[:len(path) - 1]
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def letterCombinations(self, digits: str) -> List[str]:
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if digits == "":
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return []
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self.result = []
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self._letterCombinations("", digits, 0)
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return self.result
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```
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- ### Leetcode problem 19
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> [删除链表的倒数第N个节点](https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/)
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- **算法思路：**
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- 用快慢指针法
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- 先让快指针走n步，然后让快慢指针同时右移，快指针为None时，慢指针指向的节点即为要删除的节点；
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- 对要删除的节点是头结点的情况进行特判。
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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 removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
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fast, slow, pre = head, head, self
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try:
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for i in range(n):
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fast = fast.next
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except BaseException:
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return head
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if fast is None:
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return head.next
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while fast is not None:
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fast, slow, pre = fast.next, slow.next, slow
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pre.next = slow.next
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return head
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```
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