leetCode(47):Triangle

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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.

class Solution {
public:
   int sum(vector<vector<int>>& triangle,int row,int low)
    {
        if(row==triangle.size()-1)
        {
            return triangle[row][low];
        }
        else
        {

<span style="white-space:pre">		</span>//当前值加上下一行的两个值对应的较小路径，仔细观察可知，递归的时候有重复计算，所以是可以改进的。
            return triangle[row][low]+min(sum(triangle,row+1,low),sum(triangle,row+1,low+1));
        }
    }

    int minimumTotal(vector<vector<int>>& triangle) {
        if(triangle.size()==0 || triangle[0].size()==0)
            return 0;          
        
        return sum(triangle,0,0);
    }
};

class Solution {
public:
  //本程序改变了输入的数组，如果有要求的话可以先将输入数组复杂
    int minimumTotal(vector<vector<int>>& triangle) {
        if(triangle.size()==0 || triangle[0].size()==0)
            return 0;
            
        for(int i=triangle.size()-2;i>=0;--i)
        {//从底层开始算起，本层的值等于当前值加上下一层的较小者
            for(int j=0;j<triangle[i].size();++j)
            {//这里用到了标准库内的函数
                triangle[i][j]+=min(triangle[i+1][j],triangle[i+1][j+1]);
            }
        }
        //返回第一行的第一个值即最小路径
        return triangle[0][0];
    }
};

leetCode(47):Triangle

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原文地址：http://blog.csdn.net/walker19900515/article/details/47008601