【Leetcode】120. Triangle

Description

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.

Solution

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];
    }

Analyse

这里分析了几点情况如下，其实思路就是从低往高找，思路如下所示。如果上一层是第0个，下面可以走的就是第0个和第1个，这样就是triangle.get(i).get(j)对应下面的A[j]和A[j+1]，所以要在A[j]和A[j+1]中选出一个min值，这个值用来更新。初始化的时候就是triangle.size() + 1的数组大小，因为j的范围要到triangle.get(i)里的大小加一，相当于从底部开始更新，更新到最上面，结果就是A[0]的返回。这里的复杂度是O(n^2),因为这里两重for循环。