随笔分类 - python算法
摘要:''' 1 4 3 4 5 6 5 r y b b r b b ''' def func(cards): n = len(cards) matrix = [[0 for _ in range(n)] for _ in range(n)] for i in range(n): for j in ran
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摘要:''' 有效括号: 左括号必须以正确的顺序闭合 正确【】{} () 错误 [{]}] ''' class Solution: def is_valid(self,s): dic = {'(':')','[':']','{':'}'} stack = [] for i in s: if i in di
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摘要:''' 长木板长度 longer,短木板长度 shorter,一共有k长木板,可以拼成的木板长度区间是多少? ''' class Solution: def diving_board(self,shorter,longer,k): if k == 0: return [] ans = [] minl
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摘要:''' 1 / \ 2 3 12 + 13 = 25 ''' class TreeNode: def __init__(self,val): self.val = val self.left = None self.right = None class Solution: def sum_numbe
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摘要:''' 1 / \ 2 3 / \ 4 5 ''' class TreeNode: def __init__(self,val): self.val = val self.left = None self.right = None # 深度优先-先序遍历 def dfs_pre(node): if 
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摘要:# bfs & dfs graph = { "A":["B","C"], "B":["A","C","D"], "C":["A","B","D","E"], "D":["B","C","E","F"], "E":["C&q
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摘要:基础版 青蛙一次可以跳1级或者2级台阶,问跳到N级有多少种跳法 class Solution: def jumpFloor(self,n): if n < 1: return 0 if n == 1 or n == 2: return n a = 1 b = 2 tmp = 1 for i in r
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摘要:给你一个有序列表(升序),一个目标值。写代码返回列表中和为目标值的2个数的下标(该组合唯一) input [2,7,11,15], t = 18 output 1,2 双指针法 class Solution: def two_sum(self,nums,t): left = 0 right = le
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浙公网安备 33010602011771号