标签:col res etc for def input 就是 queue 差距
1 """ 2 Given a non-negative integer numRows, generate the first numRows of Pascal‘s triangle. 3 In Pascal‘s triangle, each number is the sum of the two numbers directly above it. 4 Example: 5 Input: 5 6 Output: 7 [ 8 [1], 9 [1,1], 10 [1,2,1], 11 [1,3,3,1], 12 [1,4,6,4,1] 13 ] 14 """ 15 """ 16 解法一: 17 先来一个自己写的AC 18 直来直去,将第一行第二行单独拿出来讨论 19 第三行以后按照规律循环 20 """ 21 class Solution1: 22 def generate(self, numRows): 23 if numRows <= 0: 24 return [] 25 if numRows == 1: 26 return [[1]] 27 res = [[1], [1, 1]] 28 if numRows == 2: 29 return res 30 queue = [1, 1] 31 while numRows - 2: 32 temp = [1] 33 for i in range(len(queue)-1): 34 temp.append(queue[i] + queue[i+1]) 35 temp.append(1) 36 res.append(temp) 37 queue = temp 38 numRows -= 1 39 return res 40 """ 41 解法二:用map函数(自己知道这个函数,但想不到可以用,这就是和大佬的差距) 42 新的一行前后各添0,再相加即为下一行 43 0 1 3 3 1 44 +1 3 3 1 0 45 =1 4 6 4 1 46 """ 47 class Solution: 48 def generate(self, numRows: int) -> List[List[int]]: 49 if numRows == 0: 50 return [] 51 res = [[1]] 52 for i in range(1, numRows): 53 res.append(list(map(lambda x, y: x+y, res[-1] + [0], [0] + res[-1]))) 54 # map(function, list1, list2) 55 #res[-1]代表当前最后一行,前后添0 56 # 0 1 3 3 1 57 # +1 3 3 1 0 58 # =1 4 6 4 1 59 return res 60 # ‘+‘与extend功能相同,合二为一 []+[]=[] 61 # append是再末尾添加新的对象 [].append([]) = [[]]
标签:col res etc for def input 就是 queue 差距
原文地址:https://www.cnblogs.com/yawenw/p/12375446.html