Leet Code OJ 简单（二）

20. 有效括号 48ms

class Solution:def isValid(self, s):""":type s: str:rtype: bool"""if len(s) % 2 :return Falsebrackets = {'(': ')', '{': '}', '[': ']'}stack = []for i in s:if i in brackets:stack.append(i)else:if not stack or brackets[stack.pop()] != i:return Falseif stack:return Falsereturn True

26.删除排序数组中的重复项 96ms

class Solution:def removeDuplicates(self, nums):""":type nums: List[int]:rtype: int"""if len(nums) <= 1:return len(nums)s = 0for f in range(1, len(nums)):if nums[s] != nums[f]:s += 1nums[s] = nums[f]return s + 1

27.移除元素 56ms

class Solution:def removeElement(self, nums, val):""":type nums: List[int]:type val: int:rtype: int"""if val not in nums:return len(nums)while val in nums:nums.remove(val)return len(nums)

28.实现strStr() 48ms

class Solution:def strStr(self, haystack, needle):""":type haystack: str:type needle: str:rtype: int"""if not needle:return 0if needle not in haystack:return -1else:return haystack.index(needle)

35.搜索插入位置 48ms

class Solution:def searchInsert(self, nums, target):""":type nums: List[int]:type target: int:rtype: int"""if target in nums:return nums.index(target)if target < nums[0]:return 0if target > nums[-1]:return len(nums)for i in range(len(nums)-1):if nums[i]<target and nums[i+1]>target:return i+1

转载于:https://www.cnblogs.com/FanMLei/p/10501003.html

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