120. Triangle

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本文介绍了一种求解三角形最小路径和问题的有效算法，采用动态规划思想，利用滚动数组减少空间复杂度至O(n)，并通过具体示例进行解释。

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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.

分析：从最下面一层开始往上计算，设dp[i][j]是以第i层第j个元素为起点的最小路径和，动态规划方程如下

dp[i][j] = value[i][j] + max{dp[i-1][j], dp[i-1][j+1]}

因为每一层之和它下一层的值有关，因此只需要一个一维数组保存下层的值，代码如下：

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        const int rows = triangle.size();
        int dp[rows];
        for(int i = 0; i < rows; i++)
            dp[i] = triangle[rows-1][i];
        for(int i = rows-2; i >= 0; i--)
            for(int j = 0; j <= i; j++)
                dp[j] = triangle[i][j] + min(dp[j], dp[j+1]);
        return dp[0];
    }
};

9.14更新

--------------------

此题很有意思，用滚动数组就可以。