Pascal’s Triangle II (E)

Given a non-negative index k where k ≤ 33, return the $k^{th}$ index row of the Pascal’s triangle.

Note that the row index starts from 0.

In Pascal’s triangle, each number is the sum of the two numbers directly above it.

Example:

Input: 3
Output: [1,3,3,1]

Follow up:

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

题意

求出帕斯卡(杨辉)三角形的指定行的元素。

思路

可以直接建二维数组进行模拟；也可以压缩至一维数组进行处理；最省空间的是直接根据杨辉三角形的组合数性质直接计算出指定行的所有元素，即 $triangle[i][j]=C^j_i$ 。

代码实现 - 二维数组

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> ans = new ArrayList<>();
        int[][] triangle = new int[rowIndex + 1][rowIndex + 1];
        triangle[0][0] = 1;

        for (int i = 1; i <= rowIndex; i++) {
            for (int j = 0; j <= i; j++) {
                triangle[i][j] = j == 0 || j == i ? 1 : triangle[i - 1][j - 1] + triangle[i - 1][j];
            }
        }

        for (int i = 0; i <= rowIndex; i++) {
            ans.add(triangle[rowIndex][i]);
        }

        return ans;
    }
}

代码实现 - 一维数组

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> ans = new ArrayList<>();
        int[] row = new int[rowIndex + 1];
        row[0] = 1;

        for (int i = 1; i <= rowIndex; i++) {
            for (int j = i; j >= 1; j--) {
                row[j] = row[j] + row[j - 1];
            }
        }

        for (int i = 0; i <= rowIndex; i++) {
            ans.add(row[i]);
        }

        return ans;
    }
}

代码实现 - 一维数组(直接List处理)

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> ans = new ArrayList<>();
        ans.add(1);

        for (int i = 1; i <= rowIndex; i++) {
            for (int j = i; j >= 1; j--) {
                if (j == i) {
                    ans.add(1);
                } else {
                    ans.set(j, ans.get(j) + ans.get(j - 1));
                }
            }
        }

        return ans;
    }
}

代码实现 - 组合数

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> ans = new ArrayList<>();

        for (int i = 0; i <= rowIndex; i++) {
            ans.add(combination(rowIndex, i));
        }

        return ans;
    }

    private int combination(int i, int j) {
        if (j > i / 2) {
            return combination(i, i - j);
        }

        double ans = 1.0;
        while (j >= 1) {
            ans *= 1.0 * i-- / j--;
        }

        return (int) Math.round(ans);
    }
}

来源：https://blog.csdn.net/mapoos/article/details/100798709

标签

triangle

pascal

119. Pascal's Triangle II (E)

Pascal’s Triangle II (E)