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]
]
1 class Solution { 2 public: 3 vector<vector<int> > generate(int numRows) { 4 vector<vector<int>> sols; 5 if(numRows==0)return sols; 6 vector<int> pre(1,1); 7 sols.push_back(pre); 8 for(int i=1;i<numRows;i++) 9 { 10 vector<int> cur; 11 cur.push_back(1); 12 for(int j=1;j<i;j++) 13 { 14 cur.push_back(sols[i-1][j-1]+sols[i-1][j]); 15 } 16 cur.push_back(1); 17 sols.push_back(cur); 18 } 19 return sols; 20 } 21 };
Given an index k, return the kth row of the Pascal‘s triangle.
For example, given k = 3,
Return [1,3,3,1].
Note:
Could you optimize your algorithm to use only O(k) extra space?
思路：从后往前加就可以利用一个数组。
1 class Solution { 2 public: 3 vector<int> getRow(int rowIndex) { 4 vector<int> sols(1,1); 5 for(int i=1;i<rowIndex+1;i++) 6 { 7 sols.resize(i+1,0); 8 for(int j=sols.size()-1;j>=1;j--) 9 { 10 sols[j]=sols[j]+sols[j-1];//!!!! 11 } 12 } 13 return sols; 14 15 } 16 };
Pascal's Triangle,布布扣,bubuko.com
原文：http://www.cnblogs.com/hicandyman/p/3832818.html