给定一个非负索引 rowIndex
,返回「杨辉三角」的第 rowIndex
行。
在「杨辉三角」中,每个数是它左上方和右上方的数的和。
示例 1:
输入: rowIndex = 3 输出: [1,3,3,1]
示例 2:
输入: rowIndex = 0 输出: [1]
示例 3:
输入: rowIndex = 1 输出: [1,1]
提示:
0 <= rowIndex <= 33
进阶:
你可以优化你的算法到 O(rowIndex)
空间复杂度吗?
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
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;
}
}
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;
}
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;
}
};
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
}
impl Solution {
pub fn get_row(row_index: i32) -> Vec<i32> {
let n = (row_index + 1) as usize;
let mut res = vec![1; n];
for i in 2..n {
for j in (1..i).rev() {
res[j] += res[j - 1];
}
}
res
}
}