[LeetCode] Triangle

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.

方法一：

数组f，f[i,j] 表示从（0,0）到（i,j）的路径最小和，然后比较最后一行的最小值。top to down方法

class Solution {
    public:
    // form top to down
    int minimumTotal(vector<vector<int> > &triangle) {

        size_t rowNum = triangle.size();
        if(rowNum == 0)
            return 0;

        vector<vector<int> >f(rowNum);

        for(size_t i = 0 ; i< rowNum; i++)
        {
            f[i].resize(i+1, 0);
        }   

        for(size_t i = 0 ; i< rowNum; i++)
        {   
            //printVector(triangle[i]);
        }   

    
        f[0][0] = triangle[0][0];
        for(size_t i = 1 ; i< rowNum; i++)
        {   
            for(size_t j = 0; j <= i; j++)
            {   
                if(j == 0)
                    f[i][j] = f[i-1][j] + triangle[i][j];
                else if(j == i)
                    f[i][j] = f[i-1][j-1] + triangle[i][j];
                else
                {
                    int a = INT_MAX;
                    if(i-1 >= 0 && j-1 >= 0 && j <= i)
                        a = f[i-1][j-1];
                    int b = INT_MAX;
                    if(i-1 >= 0 && j >= 0 && j <= i-1)
                        b = f[i-1][j];
                    f[i][j] = min(a, b) + triangle[i][j];
                }
            }
        }

        int res = INT_MAX;
        for(size_t i = 0 ; i< rowNum; i++)
        {
           // printVector(f[i]);
            res = min(res, f[rowNum-1][i]);
        }

        return res;
    }
};

方法二：

bottom up方法，少了很多边界判断，f也可以只保留一行，随时更新

    int minimumTotal(vector<vector<int> > &triangle) {

        size_t rowNum = triangle.size();
        if(rowNum == 0)
            return 0;

        vector<vector<int> >f(rowNum);

        for(size_t i = 0 ; i< rowNum; i++)
        {   
            f[i].resize(i+1, 0); 
        }   

        for(size_t i = 0 ; i< rowNum; i++)
        {   
            f[rowNum-1][i] = triangle[rowNum-1][i];
        }   
            
        for(size_t i = rowNum-2 ; i >= 0; i--)
        {   
            for(size_t j = 0; j <= i; j++)
            {   
                int a = f[i+1][j];
                int b = f[i+1][j+1];
                f[i][j] = min(a, b) + triangle[i][j];
            }
        }

        return f[0][0];
    }

方法三： 在原地的数组进行运算，节省了空间

class Solution {
    public:
    int minimumTotal(vector<vector<int> > &f) {

        size_t rowNum = f.size();

            
        for(size_t i = rowNum-2 ; i >= 0; i--)
        {   
            for(size_t j = 0; j <= i; j++)
            {   

                int a = f[i+1][j];
                int b = f[i+1][j+1];
                f[i][j] = min(a, b) + f[i][j];
            }
        }

        return f[0][0];
    }
}