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]]
