Triangle
<p>Given a <code>triangle</code> array, return <em>the minimum path sum from top to bottom</em>.</p>

<p>For each step, you may move to an adjacent number of the row below. More formally, if you are on index <code>i</code> on the current row, you may move to either index <code>i</code> or index <code>i + 1</code> on the next row.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
<strong>Output:</strong> 11
<strong>Explanation:</strong> The triangle looks like:
   <u>2</u>
  <u>3</u> 4
 6 <u>5</u> 7
4 <u>1</u> 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> triangle = [[-10]]
<strong>Output:</strong> -10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= triangle.length &lt;= 200</code></li>
	<li><code>triangle[0].length == 1</code></li>
	<li><code>triangle[i].length == triangle[i - 1].length + 1</code></li>
	<li><code>-10<sup>4</sup> &lt;= triangle[i][j] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you&nbsp;do this using only <code>O(n)</code> extra space, where <code>n</code> is the total number of rows in the triangle?
Array; Dynamic Programming
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int n = triangle.size();
        int[] f = new int[n + 1];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = 0; j <= i; ++j) {
                f[j] = Math.min(f[j], f[j + 1]) + triangle.get(i).get(j);
            }
        }
        return f[0];
    }
}