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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]
]
nextline[j] = currentline[j] + currentline[j+1]
1 vector<vector<int> > generate(int numRows) 2 { 3 vector<vector<int> > pascal; 4 vector<int> line; 5 6 if (numRows <= 0) 7 return pascal; 8 9 line.push_back(1); 10 pascal.push_back(line); 11 if (numRows == 1) 12 return pascal; 13 14 for (int i = 1; i < numRows; i++) 15 { 16 vector<int> nextLine; 17 18 nextLine.push_back(1); 19 for (int j = 0; j < i - 1; j++) 20 nextLine.push_back(line[j] + line[j + 1]); 21 nextLine.push_back(1); 22 pascal.push_back(nextLine); 23 line = nextLine; 24 } 25 26 return pascal; 27 }
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?
1 vector<int> getRow(int rowIndex) 2 { 3 vector<int> line; 4 vector<int> nextline; 5 int numRows = rowIndex + 1; 6 7 if (numRows <= 0) 8 return line; 9 10 line.push_back(1); 11 if (numRows == 1) 12 return line; 13 14 for (int i = 1; i < numRows; i++) 15 { 16 nextline.push_back(1); 17 for (int j = 0; j < i - 1; j++) 18 nextline.push_back(line[j] + line[j + 1]); 19 nextline.push_back(1); 20 line = nextline; 21 nextline.clear(); 22 } 23 24 return line; 25 }
leetcode 118. Pascal's Triangle && leetcode 119. Pascal's Triangle II
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原文地址：http://www.cnblogs.com/ym65536/p/4265244.html
