John's Blogs

#加入了重复元素，会导致，无法准确的判定 mid 元素到底位于左、右哪个区域，简单的方法就是不改变顺序的去重，原来的方法即可生效Search in Rotated Sorted Array class Solution: def search(self, nums: List[int], target: int) -> int: nums = list(set(n...

class Solution: def search(self, nums: List[int], target: int) -> int: low = 0 high = len(nums)-1 while low <= high: mid = (low+high) // 2 i...

#利用字典存放匹配规则 class Solution: def isValid(self, s: str) -> bool: if s == "": return True if len(s) == 1: #只有一个字符时 return False stack = [] ...

#关键就是找到一个元素左、右边第一个比它小的元素，然后计算 宽度 × 自身高度 class Solution: def largestRectangleArea(self, heights: List[int]) -> int: ans = 0 stack = [] heights = [0] + heights + [0] #便于...

#翻转k次 等效于 从count-k那个结点开始循环遍历一遍，在k % count ==0 时，序列不变 # Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = x self.next = None class Solution...

class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: #二分查找在矩阵中的应用 if matrix == [] or matrix == [[]]: return False row = le...

class Solution: def lengthOfLastWord(self, s: str) -> int: temp=s.split(" ") for word in temp[-1::-1]: if word != "": return len(word) retu...

#利用快速排序的思想，序列两边同时进行比较，时间复杂度为：O(n) class Solution: def sortArrayByParity(self, A: List[int]) -> List[int]: i = 0 j = len(A)-1 L_num = A[j] #取出最后一个值 while i < ...

import math class Solution: def myPow(self, x: float, n: int) -> float: return math.pow(x,n) class Solution: def myPow(self, x: float, n: int) -> float: res = 1 ...

class Solution: def longestValidParentheses(self, s: str) -> int: stack = [0] ret = 0 for char in s: if (char == "("): stack.append(0) ...

当存在 [1] 这样的现象，[2] 这样的做法，是否合理以及正确，或者有 [1] 这样的现象，为什么就可以有 [2] 这样的做法 ？ 算法思想 [1] 当主串的第i个字符与子串的第j个字符失配时，若子串的前(k-1)个字符和子串的后(k-1)个字符匹配，‘p1……pk-1’=‘pj-k+1……pj-1’，则只需主串 S 的第 i 个字符与子串 P 的第 k 个字符开始向后比较即可，i 不必回溯。 ...

class Solution: def circularArrayLoop(self, nums: List[int]) -> bool: nums_len = len(nums) for i in range(nums_len): nindex = sindex = i temp =...

class Solution: def lengthOfLongestSubstring(self, s: str) -> int: longest_sub = 0 temp = [] point = 1 if len(s) <= 1: return len(s) temp...

整体框架 线性结构 线性表 非线性结构 应用

class Solution: def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float: nums = nums1 + nums2 #print(nums) nums.sort() #排序的时间为O(log(m+n))即可 ...

# Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = x self.next = None #Use listnode to get a decimal number class GetNum: snode = None ...

class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)-1): comp=target-nums[i] for j in range(i+1,len(nums)): ...