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28. Triangle && Pascal's Triangle && Pascal's Triangle II
Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[ [2], [3,4], [6,5,7], [4,1,8,3]]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
Note: Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
思想： 经典的动态规划题。
class Solution {public: int minimumTotal(vector<vector<int> > &triangle) { vector<int> pSum(triangle.size()+1, 0); for(int i = triangle.size()-1; i >= 0; --i) for(int j = 0; j < triangle[i].size(); ++j) pSum[j] = min(pSum[j]+triangle[i][j], pSum[j+1]+triangle[i][j]); return pSum[0]; }};
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]]思想： 简单的动态规划。
class Solution {public: vector<vector<int> > generate(int numRows) { vector<vector<int> > vec; if(numRows <= 0) return vec; vec.push_back(vector<int>(1, 1)); if(numRows == 1) return vec; vec.push_back(vector<int>(2, 1)); if(numRows == 2) return vec; for(int row = 2; row < numRows; ++row) { vector<int> vec2(row+1, 1); for(int Id = 1; Id < row; ++Id) vec2[Id] = vec[row-1][Id-1] + vec[row-1][Id]; vec.push_back(vec2); } return vec; }};
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].
Note: Could you optimize your algorithm to use only O(k) extra space?
思想： 动态规划。注意O(k)空间时，每次计算新的行时，要从右向左加。否则，会发生值的覆盖。
class Solution {public: vector<int> getRow(int rowIndex) { vector<int> vec(rowIndex+1, 1); for(int i = 2; i <= rowIndex; ++i) for(int j = i-1; j > 0; --j) // key, not overwrite vec[j] += vec[j-1]; return vec; }};
28. Triangle && Pascal's Triangle && Pascal's Triangle II