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129 lines
4.1 KiB
Markdown
129 lines
4.1 KiB
Markdown
# Algorithm Homework 5
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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: 56，148，274
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## 题解：
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- ### Leetcode problem 56
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> [合并区间](https://leetcode-cn.com/problems/merge-intervals/)
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- **算法思路：**
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- 首先对列表从小到大排序；
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- 然后用两个指针，一个cur指向当前值，一个next指向下一个要对比的值；
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- 如果下一个元素的左边界比当前元素的右边界要大，则无需合并，next右移；
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- 如果下一个元素的左边界小于等于当前元素的右边界，则需要合并，进一步判断下一个元素的右边界是否大于当前元素的右边界，如果是，则修改当前元素的右边界为下一个元素的右边界。接着next右移
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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 merge(self, intervals: List[List[int]]) -> List[List[int]]:
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intervals.sort()
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cur, nextElement = 0, 1
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while nextElement < len(intervals):
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if intervals[nextElement][0] <= intervals[cur][1]:
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if intervals[nextElement][1] > intervals[cur][1]:
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intervals[cur][1] = intervals[nextElement][1]
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else:
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intervals[cur + 1], cur = intervals[nextElement], cur + 1
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nextElement = nextElement + 1
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return intervals[0:cur + 1]
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```
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- ### Leetcode problem 148
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> [排序链表](https://leetcode-cn.com/problems/sort-list/)
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- **算法思路：**
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- 用快慢指针来进行将一个链表分成两个独立的子链表；
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- 采用分治和递归的思想对每个子链表进行递归排序之后归并子链表。
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- **Leetcode提交结果：**
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- **代码：**
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def sortList(self, head: ListNode) -> ListNode:
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def twoWayMerge(l1: ListNode, l2: ListNode) -> ListNode:
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if l1 is None:
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return l2
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if l2 is None:
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return l1
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head = ans = l1 if l1.val < l2.val else l2
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if l1.val < l2.val:
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l1 = l1.next
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else:
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l2 = l2.next
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while l1 is not None and l2 is not None:
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if l1.val < l2.val:
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ans.next, l1 = l1, l1.next
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else:
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ans.next, l2 = l2, l2.next
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ans = ans.next
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if l1 is not None:
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ans.next = l1
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else:
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ans.next = l2
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return head
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if head is None or head.next is None:
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return head
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fast, slow, pre = head, head, None
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while fast is not None:
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if fast.next is not None:
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slow, fast, pre = slow.next, fast.next.next, slow
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else:
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break
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if pre is not None:
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pre.next = None
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return twoWayMerge(self.sortList(head), self.sortList(slow))
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```
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- ### Leetcode problem 274
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> [H指数](https://leetcode-cn.com/problems/h-index/)
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- **算法思路：**
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- 对数组从大到小排序；
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- 则对于排序后的数组的下标为H-1个元素必然大于H，且如果存在，下标为H的元素必然小于H。
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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 hIndex(self, citations: List[int]) -> int:
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citations.sort(reverse=True)
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for i, v in enumerate(citations):
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if v < i + 1:
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return i
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return len(citations)
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```
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