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.
思路：动态规划，设发f[i,j]表示从位置(i,j)出发到达三角形底部的最小路径和，所以状态转移方程有：
f[i,j]=min(f[i+1,j],f[i+1,j+1]) (i<size-1&&i>=0,j<i+1);
参考代码：
public class Solution {
public static int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
int trianlgesize=triangle.size();
int dp[][]=new int[trianlgesize][trianlgesize];
for(int i=0;i<triangle.get(trianlgesize-1).size();i++){
dp[trianlgesize-1][i]=triangle.get(trianlgesize).get(i);
}
for(int i=trianlgesize-2;i>=0;i--){
for(int j=0;j<i+1;j++){
dp[i][j]=Math.min(dp[i+1][j], dp[i+1][j+1])+triangle.get(i).get(j);
}
}
return dp[0][0];
}
}