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‎.DS_Store‎

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‎.idea/algorithm010.iml‎

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‎.idea/inspectionProfiles/profiles_settings.xml‎

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‎.idea/modules.xml‎

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‎Week01/.idea/Week01.iml‎

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‎Week01/.idea/inspectionProfiles/profiles_settings.xml‎

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‎Week01/.idea/modules.xml‎

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‎Week01/1-两数之和.py‎

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# 给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
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#
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# 你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。
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#
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#
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#
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# 示例:
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#
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# 给定 nums = [2, 7, 11, 15], target = 9
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#
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# 因为 nums[0] + nums[1] = 2 + 7 = 9
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# 所以返回 [0, 1]
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#
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# Related Topics 数组 哈希表
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def twoSum(self, nums: List[int], target: int) -> List[int]:
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dict = {}
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for i, n in enumerate(nums):
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dict[n] = i
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for i, n in enumerate(nums):
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if target - n in dict and i != dict[target - n]:
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return [dict[target - n], i]
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# leetcode submit region end(Prohibit modification and deletion)

‎Week01/189-旋转数组.py‎

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# 给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
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#
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# 示例 1:
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#
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# 输入: [1,2,3,4,5,6,7] 和 k = 3
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# 输出: [5,6,7,1,2,3,4]
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# 解释:
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# 向右旋转 1 步: [7,1,2,3,4,5,6]
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# 向右旋转 2 步: [6,7,1,2,3,4,5]
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# 向右旋转 3 步: [5,6,7,1,2,3,4]
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#
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#
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# 示例 2:
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#
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# 输入: [-1,-100,3,99] 和 k = 2
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# 输出: [3,99,-1,-100]
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# 解释:
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# 向右旋转 1 步: [99,-1,-100,3]
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# 向右旋转 2 步: [3,99,-1,-100]
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#
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# 说明:
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#
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#
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# 尽可能想出更多的解决方案，至少有三种不同的方法可以解决这个问题。
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# 要求使用空间复杂度为 O(1) 的 原地 算法。
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#
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# Related Topics 数组
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def rotate(self, nums: List[int], k: int) -> None:
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"""
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Do not return anything, modify nums in-place instead.
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"""
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n = len(nums)
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k = k % n
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nums[:] = nums[n-k:] + nums[:n-k]
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# leetcode submit region end(Prohibit modification and deletion)

‎Week01/21-合并两个有序链表.py‎

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# 将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
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#
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#
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#
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# 示例：
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#
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# 输入：1->2->4, 1->3->4
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# 输出：1->1->2->3->4->4
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#
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# Related Topics 链表
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for singly-linked list.
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class ListNode:
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def __init__(self, val=0, next=None):
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self.val = val
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self.next = next
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class Solution:
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def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
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if not l1 or not l2:
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return l1 or l2
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if l1.val < l2.val:
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l1.next = self.mergeTwoLists(l1.next, l2)
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return l1
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else:
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l2.next = self.mergeTwoLists(l1, l2.next)
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return l2
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# leetcode submit region end(Prohibit modification and deletion)

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