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]]