Triangle(dp)

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.

代码：空间复杂度O(n²)，可以改进。

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        int row=triangle.size();
        int col=triangle[row-].size();
        int dp[row][col];
        memset(dp,,sizeof(dp));
        int sum1=;int sum2=;int index=;
        int res=(~(unsigned))>>;
        for (int i = ; i < row; ++i)
        {
            sum1+=triangle[i][];
            sum2+=triangle[i][index];
            dp[i][]=sum1;
            dp[i][index]=sum2;
            ++index;
        }
        for (int i = ; i < row; ++i){
            for (int j = ; j < triangle[i].size()-; ++j){
                dp[i][j]=min(dp[i-][j],dp[i-][j-])+triangle[i][j];
            }
        }
        for(int i=;i<col;++i){
            if(dp[row-][i]<res) res=dp[row-][i];
        }
        return res;
    }
};

改进：因为每次只需上一次的结果，直接在原处覆盖就行，空间复杂度O(n)，用temp保留本次结果，并利用上次结果dp，求完本次之后，直接用temp覆盖dp，再进行下一次。

代码：

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        if(triangle.empty())
            return ;
        int row=triangle.size();
        int col=triangle[row-].size();
        vector<int> dp(col,);
        vector<int> temp(col,);

        int sum1=;int sum2=;int index=;
        int res=(~(unsigned))>>;
        dp[]=triangle[][];

        for (int i = ; i < row; ++i){
            temp.resize(col,);
            for (int j = ; j < triangle[i].size(); ++j){
                if(j==)
                    temp[j]=dp[j]+triangle[i][j];
                else if(j==triangle[i].size()-)
                    temp[j]=dp[j-]+triangle[i][j];
                else
                    temp[j]=min(dp[j-],dp[j])+triangle[i][j];
            }
            dp=temp;
        }
        for(int i=;i<col;++i){
            if(dp[i]<res) res=dp[i];
        }
        return res;
    }
};