1 class Solution: 2 def containsDuplicate(self, nums: List[int]) -> bool: 3 #方法1:set去重,直接比较去重之后数组长度 4 if len(set(nums)) != len(nums): 5 return True 6 else: 7 return False 8 9 # 方法2:哈希表(字典) 10 numset = {} 11 for i in nums: 12 if i not in numset: 13 numset [i] = 1 14 else : 15 return True 16 return False 17 18 #方法3:排序+判断相邻元素是否重复 19 nums.sort() 20 for i in range(len(nums)-1): 21 if nums[i] == nums[i+1]: 22 return True 23 return False
1 class Solution: 2 def findTheDifference(self, s: str, t: str) -> str: 3 #方法1 4 for i in t: 5 if t.count(i) - s.count(i) == 1: 6 return i 7 8 #方法2:哈希表 9 hasmap = {} 10 11 for str in s+t: 12 if str not in hasmap: 13 hasmap[str] = 1 14 else: 15 hasmap[str] += 1 16 17 for str in hasmap: 18 if hasmap[str] % 2 != 0: 19 return str
1 class Solution: 2 def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: 3 #单调栈+(哈希映射) 4 result = [-1] * len(nums1) 5 stack = [0] #存放数组2的0号下标 6 for i in range(1,len(nums2)): 7 if nums2[i] <= nums2[stack[-1]]: 8 stack.append(i) 9 else: 10 #栈不为空再弹出元素 #s是一个持续的过程 11 while len(stack) != 0 and nums2[i] > nums2[stack[-1]]: 12 if nums2[stack[-1]] in nums1: 13 #找出对应值在数组1中的下标 14 index = nums1.index(nums2[stack[-1]]) 15 result[index] = nums2[i] 16 stack.pop() 17 stack.append(i) #找到符合条件的值之后,再将当前元素加入栈中 18 19 return result
标签:217,return,哈希,nums,元素,str,hasmap,stack,nums2 From: https://www.cnblogs.com/wuyijia/p/17662018.html