C++实现LeetCode(118.杨辉三角)
[LeetCode] 118.Pascal's Triangle 杨辉三角
Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
In Pascal's triangle, each number is the sum of the two numbers directly above it.
Example:
Input: 5
Output:
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
杨辉三角是二项式系数的一种写法,如果熟悉杨辉三角的五个性质,那么很好生成,可参见另一篇博文Pascal's Triangle II。具体生成算是:每一行的首个和结尾一个数字都是1,从第三行开始,中间的每个数字都是上一行的左右两个数字之和。代码如下:
class Solution { public: vector<vector<int>> generate(int numRows) { vector<vector<int>> res(numRows, vector<int>()); for (int i = 0; i < numRows; ++i) { res[i].resize(i + 1, 1); for (int j = 1; j < i; ++j) { res[i][j] = res[i - 1][j - 1] + res[i - 1][j]; } } return res; } };
类似题目:
Pascal's Triangle II
参考资料:
https://leetcode.com/problems/pascals-triangle/
https://leetcode.com/problems/pascals-triangle/discuss/38150/My-C%2B%2B-code-0ms
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