118. Pascal's Triangle
2017-07-26 11:51
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题目描述
Given numRows, generate the first numRows of Pascal’s triangle.For example, given numRows = 5,
Return
[ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ]
思路
很明显,杨辉三角,在欧洲称为帕斯卡三角形。找规律:
1. 每一行的第一个和最后一个是1
2. 杨辉三角的系数同时是二次项
(a+b)n
展开式中的第n行每一项的系数对应杨辉三角第n+1行每一项的系数。
3.杨辉三角中上一行相邻的两个数相加等于这一行的值。
结合2可得:
Cmn=Cm−1n−1+Cmn−1
将其展开也可验证是正确的。
代码
class Solution { public: vector< vector<int> > generate(int numRows) { vector< vector<int> > res(numRows); int i, j; for(i = 0; i < numRows; i++) { res[i].resize(i+1);//第i行分配i个列 不可一下子分配n行n列 res[i][0] = 1; res[i][res[i].size()-1] = 1; for(j = 1; j < i; j++) { res[i][j] = res[i-1][j-1] + res[i-1][j]; } } return res; } };
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