Authored by Tony Feng
Created on April 15th, 2022
Last Modified on April 15th, 2022
Task 1 - Q03. 数组中重复的数字
Question
在一个长度为 n 的数组 nums 里的所有数字都在 0～n-1 的范围内。数组中某些数字是重复的，但不知道有几个数字重复了，也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。
Solution 1
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class Solution: # Hash Map
def findRepeatNumber(self, nums: List[int]) -> int:
dict = {}
for n in nums:
if n in dict:
return n
dict[n] = 1
return -1
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Solution 2
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class Solution: # Set
def findRepeatNumber(self, nums: [int]) -> int:
s = set()
for num in nums:
if num in s:
return num
s.add(num)
return -1
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Explanation
- Solution1 and 2 use the same idea but different data structures.
- Time Complexity: O(N)
- Space Complexity: O(N)
Task 2 - Q53,I. 在排序数组中查找数字 I
Question
统计一个数字在排序数组中出现的次数。
Solution
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class Solution: # Bianry Search
'''
Find the right bound of target and target-1
'''
def search(self, nums: [int], target: int) -> int:
def findR(tar):
i, j = 0, len(nums) - 1
while i <= j:
m = i + (j - i) // 2
if nums[m] <= tar:
i = m + 1
else:
j = m - 1
return i
return findR(target) - findR(target - 1)
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Explanation
- Time Complexity: O(log N)
- Space Complexity: O(1)
Task 3 - Q53,II. 0～n-1中缺失的数字
Question
一个长度为n-1的递增排序数组中的所有数字都是唯一的，并且每个数字都在范围0～n-1之内。在范围0～n-1内的n个数字中有且只有一个数字不在该数组中，请找出这个数字。
Solution
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'''
Find the rightest index that is not equal to its element
'''
class Solution:
def missingNumber(self, nums: List[int]) -> int:
i, j = 0, len(nums)-1
while i <= j:
m = i + (j - i) // 2
if m == nums[m]:
i = m + 1
else:
j = m - 1
return i
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Explanation
- Time Complexity: O(log N)
- Space Complexity: O(1)