Algorithm-Arrays-6 Kth pascal triangle

1. 题目: Kth Row of Pascal’s Triangle

此处有关于该问题的类似解法。

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

Pascal’s triangle : To generate A[C] in row R, sum up A’[C] and A’[C-1] from previous row R - 1.

Example:

Input : k = 3

Return : [1,3,3,1]
NOTE : k is 0 based. k = 0, corresponds to the row [1].
Note:Could you optimize your algorithm to use only O(k) extra space?

输入为第k行，输出改行的元素值(k是从0开始的下标）。

2. 解题思路

1). 法1

public List<Integer> getRow(int rowIndex) {        ArrayList<Integer> result = new ArrayList<>();        if (rowIndex < 0)            return result;        result.add(1);        for (int i = 1; i <= rowIndex; i++) {            // 当前循环的result可由上一层循环的result得到            for (int j = result.size() - 2; j >= 0; j--) {                result.set(j + 1, result.get(j + 1) + result.get(j));            }            result.add(1);        }        return result;    }

2). 法2

/*     * C(line, i) = line! / ( (line-i)! * i! ) C(line, i-1) = line! / ( (line -     * i + 1)! * (i-1)! ) We can derive following expression from above two     * expressions. C(line, i) = C(line, i-1) * (line - i + 1) / i     *      * So C(line, i) can be calculated from C(line, i-1) in O(1) time     */    public ArrayList<Integer> getRow1(int A) {        ArrayList<Integer> list = new ArrayList<Integer>(A + 1);        int line = A + 1;        int C = 1; // used to represent C(line, i)        for (int i = 1; i <= line; i++) {            list.add(C); // The first value in a line is always 1            C = C * (line - i) / i;        }        return list;    }