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).
若不考虑空间复杂度可以从上到下进行动态规划求值 再对最后一行求最小值 但空间复杂度为O(n^2) 要想空间复杂度为O(n),要从下到上进行 对于每任以层的任一点 其最小值为其下一层对应两点的最小值加其本身值 根据此可以得出代码如下:
public int minimumTotal(List<List<Integer>> triangle) { int n=triangle.size(); int[] res=new int[n+1]; for(int i=n-1;i>=0;i--){ for(int j=0;j<i+1;j++){ res[j]=triangle.get(i).get(j)+Math.min(res[j],res[j+1]); } } return res[0]; }
原文:http://blog.csdn.net/u012734829/article/details/42394391