# 【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.
> 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.

> 給一三角形，找到一條從上到下最小總和的路徑，回傳總和值。
> 注意：
> 額外的挑戰：你只能使用O(n)的空間複雜度，n代表三角形的高度。

## Example:

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).

## Solution

* 從三角形的底部往上跑，讓每個數字去加自己下面兩個數字之中比較小的。一路加到頂端，頂端就是答案。

### Code

class Solution {
public:    
    int minimumTotal(vector<vector<int>>& triangle) {
        for(int i=triangle.size()-2;i>=0;i--)
        {
            for(int j=0;j<triangle[i].size();j++)
            {
                triangle[i][j]+=min(triangle[i+1][j],triangle[i+1][j+1]);
            }
        }
        return triangle[0][0];
    }
};

###### tags: `LeetCode` `C++`