LeetCode-动态规划之Triangle

Posted 2020-10-06

题目描述

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.

分析

1. 从下到上。
2. 我们从底行[size（）- 2]上方的行开始。
3. 每个数在其下面加上两个数字较小的数字。
4. 最后我们到最上面我们是最小的总和。

Java代码

import java.util.List;

public class Triangle {

    public int minimumTotal(List<List<Integer>> triangle) {
        int[] A = new int[triangle.size() + 1];
        for (int i = triangle.size() - 1; i >= 0; i--) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                A[j] = Math.min(A[j], A[j + 1]) + triangle.get(i).get(j);
            }
        }

        return A[0];
    }
}

测试代码：

package dynamicPlanning;

import org.junit.Test;

import java.util.ArrayList;
import java.util.List;

/**
 * Created by Feng on 2017/9/11.
 */
public class TriangleTest {

    Triangle triangle = new Triangle();

    @Test
    public void minimumTotal() throws Exception {

        List<Integer> list1 = new ArrayList<>();
        list1.add(2);

        List<Integer> list2 = new ArrayList<>();
        list2.add(3);
        list2.add(4);

        List<Integer> list3 = new ArrayList<>();
        list3.add(6);
        list3.add(5);
        list3.add(7);

        List<Integer> list4 = new ArrayList<>();
        list4.add(4);
        list4.add(1);
        list4.add(8);
        list4.add(3);

        List<List<Integer>> lists = new ArrayList<>();

        lists.add(list1);
        lists.add(list2);
        lists.add(list3);
        lists.add(list4);

        int result = triangle.minimumTotal(lists);
        //System.out.println(lists.size());
        System.out.println(result);
    }

}