动态规划—triangle

题目：

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 is11(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：

题目的意思是找出从顶到底的最小路径和，类似二叉树。

从下向上进行，例如第i+1行的第j和j+1元素进行比较，将两元素中较小的值与第i行的第j个元素相加，以此类推，最后顶元素就是要求的最小路径和。

 

代码：

public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
        int row = triangle.size();
        for(int i=row-2;i>=0;i--){
            for(int j=0;j<=i;j++){
                int min = Math.min(triangle.get(i+1).get(j), triangle.get(i+1).get(j+1));
                triangle.get(i).set(j, triangle.get(i).get(j)+min);
            }
        }
        return triangle.get(0).get(0);
}