README_EN.md

119. Pascal's Triangle II

中文文档

Description

Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it as shown:

 

Example 1:

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

Example 2:

Input: rowIndex = 0
Output: [1]

Example 3:

Input: rowIndex = 1
Output: [1,1]

 

Constraints:

• 0 <= rowIndex <= 33

 

Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Solutions

Python3

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        row = [1] * (rowIndex + 1)
        for i in range(2, rowIndex + 1):
            for j in range(i - 1, 0, -1):
                row[j] += row[j - 1]
        return row

Java

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> row = new ArrayList<>();
        for (int i = 0; i < rowIndex + 1; ++i) {
            row.add(1);
        }
        for (int i = 2; i < rowIndex + 1; ++i) {
            for (int j = i - 1; j > 0; --j) {
                row.set(j, row.get(j) + row.get(j - 1));
            }
        }
        return row;
    }
}

TypeScript

function getRow(rowIndex: number): number[] {
    let ans = new Array(rowIndex + 1).fill(1);
    for (let i = 2; i < rowIndex + 1; ++i) {
        for (let j = i -1; j > 0; --j) {
            ans[j] += ans[j - 1];
        }
    }
    return ans;
};

C++

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> row(rowIndex + 1, 1);
        for (int i = 2; i < rowIndex + 1; ++i) {
            for (int j = i - 1; j > 0; --j) {
                row[j] += row[j - 1];
            }
        }
        return row;
    }
};

Go

func getRow(rowIndex int) []int {
	row := make([]int, rowIndex+1)
	row[0] = 1
	for i := 1; i <= rowIndex; i++ {
		for j := i; j > 0; j-- {
			row[j] += row[j-1]
		}
	}
	return row
}

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