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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) { //空间复杂度为O(n)，n为三角形的层数，时间复杂度为O(K)，K为整个三角形中数字的个数 //从底向上计算，动态规划，题意是只能取相邻的，因此状态转移方程可以求得 int n=triangle.size(); int dp[]=new int[n]; for(int i=n-1;i>=0;i--){ for(int j=0;j<=i;j++){ if(i==n-1){ dp[j]=triangle.get(i).get(j); }else{ dp[j]=Math.min(dp[j],dp[j+1])+triangle.get(i).get(j); } } } return dp[0]; } }
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原文地址：http://www.cnblogs.com/qiaomu/p/4668925.html
