118. Pascal's Triangle&&119. Pascal's Triangle II
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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]]
解法:没有什么,用一个成员是vector的vector逐层生成。
class Solution {public: vector<vector<int>> generate(int numRows) { vector<vector<int>> r; vector<int> pre; vector<int> p; for(int i=0;i<numRows;i++){ p.clear(); for(int j=0;j<=i;j++){ if(j==i||j==0)p.push_back(1); else p.push_back(pre[j-1]+pre[j]); } r.push_back(p); pre=p; } return r; }};
这道题更深一层就是 Pascal's Triangle II
Given an index k, return the kth row of the Pascal's triangle.
For example, given k = 3,
Return [1,3,3,1]
.
解法:有一种是跟着上一道题,直接返回最后一层。但是这样……效率……
我就知道这种题一定要找规律……我还是太嫩了,最后找到有一个很厉害的公式
v[i]=v[i-1]*(rowIndex+1-i)/i;
class Solution {public: vector<int> getRow(int rowIndex) { vector<int> v; v.resize(rowIndex+1); v[0]=v[rowIndex]=1; for(int i=1;i<=v.size()/2;i++){ v[i]=v[rowIndex-i]=v[i-1]*(rowIndex+1-i)/i; } return v; }};
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