pascal三角原理+zip用法-【leetcode119-pascal triangle2】

一、题目

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

Note:
Could you optimize your algorithm to use only O(k) extra space?

二、渣渣方法一：与118题一样，区别是只输出第k行，占用空间大

class Solution(object):    def getRow(self, rowIndex):        """        :type rowIndex: int        :rtype: List[int]        """        array = []         for i in range(rowIndex+1):        array.append([1]*(i+1))         for j in range(i):        array[i][j] = array[i-1][j-1] + array[i-1][j]        array[i][0] = array[i][-1] = 1        return array[rowIndex]

三、用pascal原理，注意zip的使用

class Solution(object):    def getRow(self, rowIndex):        """        :type rowIndex: int        :rtype: List[int]        """        row = [1]        for i in range(rowIndex):        row = [x + y for x, y in zip([0]+row,row+[0])]        return row

四、pascal三角原理

Say we have the current layer [1, 2, 1]. We then make 2 copies of this layer, add 0 to the start of one copy, and add 0 to the end of one copy; then we have [0, 1, 2, 1] and [1, 2, 1, 0]. Then we can perform the element-wise add operation and we would have [1, 3, 3, 1]. This is from the definition of Pascal's Triangle.

五、zip用法

 a = [1,2,3] b = [4,5,6] c = [4,5,6,7,8]zipped = zip(a,b)>>>[(1, 4), (2, 5), (3, 6)]zip(a,c)>>>[(1, 4), (2, 5), (3, 6)]zip(*zipped)>>>[(1, 2, 3), (4, 5, 6)]

 a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]zip(*a)>>>[(1, 4, 7), (2, 5, 8), (3, 6, 9)]map(list,zip(*a))>>>[[1, 4, 7], [2, 5, 8], [3, 6, 9]]