leetcode-Triangle

Difficulty: Medium

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.

第一种方法，递归，超时

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

第二种方法，通过，但空间复杂度不理想

class Solution {  
    public:  
        int minimumTotal(vector<vector<int> > &triangle) {  
            int n=triangle.size();  
            int m=triangle[n-1].size();
            if(n<1) return 0;  
              
            int b[n][m];
            for(int i=0;i<m;++i)
                b[n-1][i]=triangle[n-1][i];
              
            for(int i=n-2; i>=0; i--) 
                for(int j=0; j<triangle[i+1].size(); j++)
                    b[i][j] = triangle[i][j] + min(b[i+1][j], b[i+1][j+1]);  
                   
            return b[0][0];  
        }  
    };  

第三种优化

    class Solution {  
    public:  
        int minimumTotal(vector<vector<int> > &triangle) {  
            int sz = triangle.size();  
            if(sz<1) return 0;  
              
            vector<int> b = triangle[sz-1];  
              
            for(int i=sz-2; i>=0; --i) {  
                for(int j=0; j<triangle[i].size(); ++j) {  
                    b[j] = triangle[i][j] + min(b[j], b[j+1]);  
                }   
            }  
            return b[0];  
        }  
    };