leetcode_118_Pascal's Triangle
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Pascal's Triangle
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]
]
//vs2012测试代码 #include<iostream>#include<vector>using namespace std;vector<vector<int> > generate(int numRows) {vector<vector<int>> ans;//resize是改变容器的大小,并且创建对象; reserve是容器预留空间,但并不真正创建元素对象//使用assign()将10个整数值42的副本赋值给vector: v.assign( 10, 42 );ans.resize(numRows); for(int i=0; i<numRows; i++)ans[i].assign(i+1,1);for(int i=2; i<numRows; i++)for(int j=1; j<ans[i].size()-1; j++)ans[i][j] = ans[i-1][j-1] + ans[i-1][j];for(int i=0; i<numRows; i++){for(int j=0; j<ans[i].size(); j++)cout<<ans[i][j];cout<<endl;}return ans;}int main(){int num;cin>>num;generate(num);return 0;}
//方法一:自测Acceptedclass Solution {public: vector<vector<int> > generate(int numRows) {vector<vector<int>> ans;//resize是改变容器的大小,并且创建对象; reserve是容器预留空间,但并不真正创建元素对象//使用assign()将10个整数值42的副本赋值给vector: v.assign( 10, 42 );ans.resize(numRows); for(int i=0; i<numRows; i++)ans[i].assign(i+1,1);for(int i=2; i<numRows; i++)for(int j=1; j<ans[i].size()-1; j++)ans[i][j] = ans[i-1][j-1] + ans[i-1][j];return ans; }};
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