| comments | difficulty | edit_url | tags | ||
|---|---|---|---|---|---|
true |
简单 |
|
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 fclass 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 = 1; j < f.get(i).size(); ++j) {
g.add(f.get(i).get(j - 1) + f.get(i).get(j));
}
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 = 1; j < f[i].size(); ++j) {
g.push_back(f[i][j - 1] + f[i][j]);
}
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 := 1; j < len(f[i]); j++ {
g = append(g, f[i][j-1]+f[i][j])
}
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 = 1; j < f[i].length; ++j) {
g.push(f[i][j - 1] + f[i][j]);
}
g.push(1);
f.push(g);
}
return f;
}impl Solution {
pub fn generate(num_rows: i32) -> Vec<Vec<i32>> {
let mut f = vec![vec![1]];
for i in 1..num_rows {
let mut g = vec![1];
for j in 1..f[i as usize - 1].len() {
g.push(f[i as usize - 1][j - 1] + f[i as usize - 1][j]);
}
g.push(1);
f.push(g);
}
f
}
}/**
* @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 = 1; j < f[i].length; ++j) {
g.push(f[i][j - 1] + f[i][j]);
}
g.push(1);
f.push(g);
}
return f;
};public class Solution {
public IList<IList<int>> Generate(int numRows) {
var f = new List<IList<int>> { new List<int> { 1 } };
for (int i = 1; i < numRows; ++i) {
var g = new List<int> { 1 };
for (int j = 1; j < f[i - 1].Count; ++j) {
g.Add(f[i - 1][j - 1] + f[i - 1][j]);
}
g.Add(1);
f.Add(g);
}
return f;
}
}