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-1].size();
        int dp[row][col];
        memset(dp,0,sizeof(dp));
        int sum1=0;int sum2=0;int index=0;
        int res=(~(unsigned)1)>>1;
        for (int i = 0; i < row; ++i)
        {
            sum1+=triangle[i][0];
            sum2+=triangle[i][index];
            dp[i][0]=sum1;
            dp[i][index]=sum2;
            ++index;
        }
        for (int i = 1; i < row; ++i){
            for (int j = 1; j < triangle[i].size()-1; ++j){
                dp[i][j]=min(dp[i-1][j],dp[i-1][j-1])+triangle[i][j];
            }
        }
        for(int i=0;i<col;++i){
            if(dp[row-1][i]<res) res=dp[row-1][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 0;
        int row=triangle.size();
        int col=triangle[row-1].size();
        vector<int> dp(col,0);
        vector<int> temp(col,0);

        int sum1=0;int sum2=0;int index=0;
        int res=(~(unsigned)1)>>1;
        dp[0]=triangle[0][0];

        for (int i = 1; i < row; ++i){
            temp.resize(col,0);
            for (int j = 0; j < triangle[i].size(); ++j){
                if(j==0) 
                    temp[j]=dp[j]+triangle[i][j];
                else if(j==triangle[i].size()-1)
                    temp[j]=dp[j-1]+triangle[i][j];
                else 
                    temp[j]=min(dp[j-1],dp[j])+triangle[i][j];
            }
            dp=temp;
        }
        for(int i=0;i<col;++i){
            if(dp[i]<res) res=dp[i];
        }
        return res; 
    }
};