标签：style blog color io strong for ar div log
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];
}
};
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;
}
};
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
标签：style blog color io strong for ar div log
原文地址：http://www.cnblogs.com/liyangguang1988/p/3939471.html
