[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
;
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];
}
}