Set Dict
- 219. 存在重复元素 II
- 136. 只出现一次的数字
- 846. 一手顺子
219. 存在重复元素 II
class Solution:
def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
n = len(nums)
# 1、双 for 超时
# for i in range(n):
# for j in range(max(0,i-k),i):
# if nums[i] == nums[j]:
# return True
# 2、in
# for i in range(n):
# if nums[i] in set(nums[i+1:i+1+k]):return True
# 3、set
s, c = set(), 0
for i, num in enumerate(nums):
if num in s: return True
s.add(num)
# 1) 计数
# c += 1
# if c > k:
# s.remove(nums[i - k])
# c -= 1
# 2) 计数
# if c < k: c += 1
# else: s.remove(nums[i - k])
# 3) len
if len(s) > k: s.remove(nums[i - k])
return False136. 只出现一次的数字
class Solution:
def singleNumber(self, nums: List[int]) -> int:
## 方法一：dict
d = {}
for i in nums: d[i] = d.get(i, 0) + 1
return [x for x in d if d[x] == 1][0]
## 方法二：Counter
count = Counter(nums)
for i in count:
if count[i] == 1: return i
## 方法五：
return sum(set(nums))*2-sum(nums)846. 一手顺子
class Solution:
def isNStraightHand(self, hand: List[int], groupSize: int) -> bool:
if len(hand) % groupSize: return False
hand.sort()
cnt = Counter(hand)
for h in hand:
if cnt[h] == 0: continue
for num in range(h, h + groupSize):
if cnt[num] == 0: return False
cnt[num] -= 1
return True
