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简单
数组
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English Version

题目描述

给定一个非负整数 numRows生成「杨辉三角」的前 numRows 行。

「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

解法

方法一:模拟

我们先创建一个答案数组 $f$,然后将 $f$ 的第一行元素设为 $[1]$。接下来,我们从第二行开始,每一行的开头和结尾元素都是 $1$,其它 $f[i][j] = f[i - 1][j - 1] + f[i - 1][j]$

时间复杂度 $O(n^2)$,其中 $n$ 为给定的行数。忽略答案的空间消耗,空间复杂度 $O(1)$

Python3

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

Java

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 = 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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

JavaScript

/**
 * @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;
};

C#

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;
    }
}