Given a triangle array, return the minimum path sum from top to bottom.
For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.
\n
Example 1:
\n\n\nInput: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]\nOutput: 11\nExplanation: The triangle looks like:\n 2\n 3 4\n 6 5 7\n4 1 8 3\nThe minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).\n\n\n
Example 2:
\n\n\nInput: triangle = [[-10]]\nOutput: -10\n\n\n
\n
Constraints:
\n\n1 <= triangle.length <= 200triangle[0].length == 1triangle[i].length == triangle[i - 1].length + 1-104 <= triangle[i][j] <= 104\nFollow up: Could you do this using only
O(n) extra space, where n is the total number of rows in the triangle?",
"translatedTitle": "三角形最小路径和",
"translatedContent": "给定一个三角形 triangle ，找出自顶向下的最小路径和。
每一步只能移动到下一行中相邻的结点上。相邻的结点 在这里指的是 下标 与 上一层结点下标 相同或者等于 上一层结点下标 + 1 的两个结点。也就是说，如果正位于当前行的下标 i ，那么下一步可以移动到下一行的下标 i 或 i + 1 。
\n\n
示例 1：
\n\n\n输入：triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]\n输出：11\n解释：如下面简图所示：\n 2\n 3 4\n 6 5 7\n4 1 8 3\n自顶向下的最小路径和为 11（即，2 + 3 + 5 + 1 = 11）。\n\n\n
示例 2：
\n\n\n输入：triangle = [[-10]]\n输出：-10\n\n\n
\n\n
提示：
\n\n1 <= triangle.length <= 200triangle[0].length == 1triangle[i].length == triangle[i - 1].length + 1-104 <= triangle[i][j] <= 104\n\n
进阶：
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