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.
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.
很明显的动态规划,当前总值只和上面的左右有关系,一次遍历就搞定
class Solution {
public:
int minimumTotal(vector<vector<int>>& tri)
{
for(int i=1;i<tri.size();i++)
{
tri[i][0] += tri[i-1][0];
tri[i][i] += tri[i-1][i-1];
for (int j = 1; j < i; j++)
tri[i][j] += min(tri[i-1][j-1], tri[i-1][j]);
}
return *min_element(tri[tri.size() - 1].begin(), tri[tri.size() - 1].end());
}
};