119. Pascal's Triangle II——array

题目分析：因为只能用k的空间，所以考虑递归。但是递归超过了时间，所以考虑特性，每个值是前liang两个值得和，倒序计算。

class Solution(object):
    def kthSmallest(self, rowIndex):
    #     res = [0 for _ in range(rowIndex+1)]
    #     res[0] = 1
    #     res[-1] = 1
    #     for i in range(1, (rowIndex+1)//2+(rowIndex+1)%2):
    #         res[i] = self.s(rowIndex, i)
    #     for j in range((rowIndex+1)//2+(rowIndex+1)%2, rowIndex+1):
    #         res[j] = res[rowIndex-j]
    #     return res
    #
    # def s(self, row, colum):
    #     if colum == 0 or row == colum:
    #         return 1
    #     else:
    #         return self.s(row-1, colum-1)+self.s(row-1, colum)


        result = [1]
        if rowIndex == 0:
            return result
        result.append(1)
        if rowIndex == 1:
            return result
        for i in range(2, rowIndex + 1):
            result.append(1)
            n = len(result)
            # 逆序的原因是防止前面变了影响后面
            for j in range(n - 2, 0, -1):
                if (j >= n // 2):
                    result[j] = result[j] + result[j - 1]
                else:
                    result[j] = result[n - 1 - j]
        return result