【leetcode】Pascal's Triangle I & II (middle)
2015-04-10 17:01
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Given numRows, generate the first numRows of Pascal's triangle.
For example, given numRows = 5,
Return
思路:杨辉三角,直接按规律生成即可
Given an index k, return the kth row of the Pascal's triangle.
For example, given k = 3,
Return
Note:
Could you optimize your algorithm to use only O(k) extra space?
思路:
要靠数学公式了,设杨辉三角的最顶层为第0行,每行的第一个数字是第0个,则
第 i 行第 j 个元素的计算为:C(i, j) = (i)! / (j)! * (i - j)!
那么第 i 行第 j 个元素和它前一个元素的关系是 ans[i][j] = ans[i][j - 1] * (i - j + 1) / j; //这里要注意不要越界
For example, given numRows = 5,
Return
[ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ]
思路:杨辉三角,直接按规律生成即可
vector<vector<int> > generate(int numRows) { vector<vector<int>> ans; for(int i = 0; i < numRows; i++) { vector<int> v(i + 1, 1); for(int j = 1; j < i; j++) { v[j] = ans[i - 1][j - 1] + ans[i - 1][j]; } ans.push_back(v); } return ans; }
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:
要靠数学公式了,设杨辉三角的最顶层为第0行,每行的第一个数字是第0个,则
第 i 行第 j 个元素的计算为:C(i, j) = (i)! / (j)! * (i - j)!
那么第 i 行第 j 个元素和它前一个元素的关系是 ans[i][j] = ans[i][j - 1] * (i - j + 1) / j; //这里要注意不要越界
vector<int> getRow(int rowIndex) { vector<int> ans(rowIndex + 1, 1); ans[rowIndex - 1] = ans[1] = rowIndex; for(int i = 2; i < rowIndex / 2 + 1; i++) { ans[rowIndex - i] = ans[i] = (long long)ans[i - 1] * (rowIndex - i + 1) / i; //用long long防止越界 同时利用杨辉三角的对称性减少一半的计算量 } return ans; }
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