【Python成长之路】Leetcode 456题：132模式解法汇总

class Solution(object):    def find132pattern(self, nums):        if len(nums) < 3:            return False        for i in range(len(nums)):            first = nums[i]            for j in range(i + 1, len(nums)):                if nums[j] > first:                    mid = nums[j]                    for k in range(j+1,len(nums)):                        if first < nums[k] < mid:                            return True        return False

import itertoolsclass Solution(object):    def find132pattern(self, nums):        if len(nums) < 3:            return False        res = itertools.combinations(nums, 3)        for i in res:            if i[0] < i[2] < i[1]:                return True        return False

class Solution:    def find132pattern(self, nums: List[int]) -> bool:        stack = []        for num in nums:            if len(stack) == 0 or num < stack[-1][0]:                stack.append([num, num])            else:                top_min = stack[-1][0]                while len(stack) != 0 and num > stack[-1][1]:                    stack.pop()                    if len(stack) != 0 and num > stack[-1][0] and num < stack[-1][1]:                        return True                    stack.append([top_min, num])                return False              

class Solution:    def find132pattern(self, nums: List[int]) -> bool:        stack=[]        _Min=float("-inf")        for i in range(len(nums)-1,-1,-1):            if nums[i]<_Min:                return True            while stack and nums[i]>stack[-1]:                _Min=stack.pop()            stack.append(nums[i])        return False

1、带索引、值的遍历（这个应该很常见）for i,k in enumerate(mylist)
2、求列表排序后，之前的索引值列表[i for i,k in sorted(enumerate(mylist),reverse=True,key=lambda x:x[1])]
3、求列表某值的索引位置[i for i,k in enumerate(mylist) if k == num]
4、三元表达式return True if x>y else False
5、将字符串大小写进行转换mystr.swapcase()
6、将列表中的字符串按数字进行排序，如 ['b3','a10','c1']return sorted(mylist,key=lambda x:int(re.search('\d+').group(0)))
7、对两个列表进行元素交接，使两列表的和差值最小list3 = list2 + list1list1 = min(itertools.combinations(list3,len(list1)),key=lambda x:abs(sum(x)-sum(list3)/2))
8、求列表各元素的频度return {i:mylist.count(k) for i,k in enumerate(set(mylist))}