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