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.
hint : 简单的动态规划，若考虑 状态 dp[i][j] := 从开始位置到第i行j列的最小路径和，那么可以给出转移方程为：dp[i][j] = triangle[i][j] + min{dp[i-1][j],dp[i-1][j-1]}, for( 0
public class Solution {
public int minimumTotal(ArrayList> triangle) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if(triangle.size() == 0)
return 0;
if(triangle.size() == 1)
return triangle.get(0).get(0);
int n = triangle.size();
int[] dplast = new int[1];
dplast[0] = triangle.get(0).get(0);
for(int i = 1; i < n; i++)
{
ArrayList
tmp = triangle.get(i);
int[] dpcurrent = new int[tmp.size()];
for(int j = 0; j < tmp.size(); j++)
{
if(j == 0)
{
dpcurrent[j] = dplast[j] + tmp.get(j);
continue;
}
if(j == tmp.size() - 1)
{
dpcurrent[j] = dplast[j-1] + tmp.get(j);
continue;
}
dpcurrent[j] = Math.min(dplast[j-1], dplast[j]) + tmp.get(j);
}
dplast = dpcurrent;
}
int res = Integer.MAX_VALUE;
for(int i = 0; i < n; i++)
{
res = Math.min(res, dplast[i]);
}
dplast = null;
return res;
}
}