（一）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.
- 自底向上，从倒数第二行开始
- 每个元素 = 当前元素 + min（下一行相邻第一个元素，下一行相邻第二个元素）
- 遍历之后，第一行第一个元素就是结果
- 最优子结构：F[i][j] = F[i][j] + min(F[i+1][j],F[i+1][j+1])
package DynamicProgramming.Triangle;
import java.util.ArrayList;
public class Solution {
public static int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
int size = triangle.size(); //三角形的行数
for(int i= size - 2; i >= 0; i--){
for(int j=0; j <= i; j++){
triangle.get(i).set(j,triangle.get(i).get(j) + Math.min(triangle.get(i+1).get(j),triangle.get(i+1).get(j+1)));
}
}
return triangle.get(0).get(0);
}
public static void main(String[] args){
ArrayList<ArrayList<Integer>> triangle = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> l1 = new ArrayList<Integer>();
ArrayList<Integer> l2 = new ArrayList<Integer>();
ArrayList<Integer> l3 = new ArrayList<Integer>();
ArrayList<Integer> l4 = new ArrayList<Integer>();
l1.add(2);
l2.add(3);
l2.add(4);
l3.add(6);
l3.add(5);
l3.add(7);
l4.add(4);
l4.add(1);
l4.add(8);
l4.add(3);
triangle.add(l1);
triangle.add(l2);
triangle.add(l3);
triangle.add(l4);
System.out.println(minimumTotal(triangle));
}
}