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.

思路：
采用DP算法。本题可以采用自顶向下和自底向上2种方式，但是由于题目要求只使用(O(n))的额外空间，自顶向下需要(O(n^2))的额外空间，而自底向上通过重复上一层覆盖下一层的值可以做到空间复杂度为(O(n))，所以选择自底向上的方式。每个父节点只需要选择子节点中最小的路径再加上自己的值即成为到达父节点的最短路径。更新的公式如下：min_path[i] = min(min_path[i], min_path[i+1]) + triangle[layer[i]。

Solution (C++):

int minimumTotal(vector<vector<int>>& triangle) {
    int n = triangle.size();
    vector<int> min_path(triangle.back());    //使用最底层的值初始化
    
    for (int layer = n-2; layer >= 0; --layer) {
        for (int i = 0; i <= layer; i++) {
            min_path[i] = min(min_path[i], min_path[i+1]) + triangle[layer][i];    //选取路径最短的子节点
        }
    }
    return min_path[0];
}

性能：

Runtime: 8 ms  Memory Usage: 9.8 MB