[leedcode 120] Triangle

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];
    }
}