leetcode || 120、Triangle

problem：

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.

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Array Dynamic
Programming

题意：给一个三角形矩阵（左下型），从上到下，每次遍历两行之间的相邻元素，求最小路径和

thinking：

考察DP，开一个n(三角型矩阵的最后一行的大小)大小的数组，每次记录上一行到下一行以该元素为终点的路径和，

这样，最后在数组中寻找最小的一个元素就是最小路径和

code：

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

        int ret = INT_MAX;
        for(int i = 0; i < f.size(); i++)
            ret = min(ret, f[i]);

        return ret;
    }
};