给定一个非负整数 numRows
,生成「杨辉三角」的前 numRows
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: numRows = 5 输出: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
示例 2:
输入: numRows = 1 输出: [[1]]
提示:
1 <= numRows <= 30
我们先创建一个答案数组
时间复杂度
class Solution:
def generate(self, numRows: int) -> List[List[int]]:
f = [[1]]
for i in range(numRows - 1):
g = [1] + [a + b for a, b in pairwise(f[-1])] + [1]
f.append(g)
return f
class Solution {
public List<List<Integer>> generate(int numRows) {
List<List<Integer>> f = new ArrayList<>();
f.add(List.of(1));
for (int i = 0; i < numRows - 1; ++i) {
List<Integer> g = new ArrayList<>();
g.add(1);
for (int j = 0; j < f.get(i).size() - 1; ++j) {
g.add(f.get(i).get(j) + f.get(i).get(j + 1));
}
g.add(1);
f.add(g);
}
return f;
}
}
class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<vector<int>> f;
f.push_back(vector<int>(1, 1));
for (int i = 0; i < numRows - 1; ++i) {
vector<int> g;
g.push_back(1);
for (int j = 0; j < f[i].size() - 1; ++j) {
g.push_back(f[i][j] + f[i][j + 1]);
}
g.push_back(1);
f.push_back(g);
}
return f;
}
};
func generate(numRows int) [][]int {
f := [][]int{[]int{1}}
for i := 0; i < numRows-1; i++ {
g := []int{1}
for j := 0; j < len(f[i])-1; j++ {
g = append(g, f[i][j]+f[i][j+1])
}
g = append(g, 1)
f = append(f, g)
}
return f
}
function generate(numRows: number): number[][] {
const f: number[][] = [[1]];
for (let i = 0; i < numRows - 1; ++i) {
const g: number[] = [1];
for (let j = 0; j < f[i].length - 1; ++j) {
g.push(f[i][j] + f[i][j + 1]);
}
g.push(1);
f.push(g);
}
return f;
}
impl Solution {
#[allow(dead_code)]
pub fn generate(num_rows: i32) -> Vec<Vec<i32>> {
let mut ret_vec: Vec<Vec<i32>> = Vec::new();
let mut cur_vec: Vec<i32> = Vec::new();
for i in 0..num_rows as usize {
let mut new_vec: Vec<i32> = vec![1; i + 1];
for j in 1..i {
new_vec[j] = cur_vec[j - 1] + cur_vec[j];
}
cur_vec = new_vec.clone();
ret_vec.push(new_vec);
}
ret_vec
}
}
/**
* @param {number} numRows
* @return {number[][]}
*/
var generate = function (numRows) {
const f = [[1]];
for (let i = 0; i < numRows - 1; ++i) {
const g = [1];
for (let j = 0; j < f[i].length - 1; ++j) {
g.push(f[i][j] + f[i][j + 1]);
}
g.push(1);
f.push(g);
}
return f;
};