leetcode[120] Triangle

给定一个三角数组，找从上到下的最小和，例如：

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).

思路一，DP， 在原三角数组中操作，每层更新为到该层的最小距离，然后返回最后一层的最小即可。

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

思路二，不改变原来数组，类似上一题，另建空间O(row)来存到当前行的最优，然后根据数组来更新这个最优数组，然后在数组中返回最小的值。

class Solution {
public:
int minimumTotal(vector<vector<int> > &triangle)
{
    int ans = INT_MAX, row = triangle.size(), pre;
    if (row == 0) return ans;
    vector<int> tmp;
    tmp.push_back(triangle[0][0]);
    for (int i = 1; i < row; ++i)
    {
        pre = tmp[0];
        tmp[0] += triangle[i][0];
        for (int j = 1; j < tmp.size(); ++j)
        {
            int mid = tmp[j];
            tmp[j] = min(pre, tmp[j]) + triangle[i][j];
            pre = mid;
        }
        tmp.push_back(pre + triangle[i][triangle[i].size()-1]);
    }
    for (int i = 0; i < tmp.size(); ++i)
    {
        if (tmp[i] < ans)
            ans = tmp[i];
    }
    return ans;
}
};

耗时一样。空间不一样。