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.
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int n = triangle.size();
if(n==1)
return triangle.get(n-1).get(n-1);
int[][] f = new int[n][n];
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
f[i][j] = 0;
}
for(int j=0;j<n;j++)
f[n-1][j] = triangle.get(n-1).get(j); // 需要把数组最后一行的值赋给f[n-1][j]
for(int i=n-2;i>=0;i--)
for(int j=0;j<=i;j++)
{
f[i][j] = Math.min(f[i+1][j],f[i+1][j+1]) + triangle.get(i).get(j); // f(i,j) = min{f(i+1,j),f(i+1,j+1)} + triangle(i,j);
}
return f[0][0];
}
}
