Triangle leetcode

 

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.

 

 

解题思路：使用动态规划法。当我们计算第i层的数到底层的最小和时，如果我们知道第i+1层的数到底层最小的和就好算了。即minsum[i][j]=triangle[i]+min( minsum[i+1][j] , minsum[i+1][j+1] )；从底层向顶层逐层计算，就能得到最终结果。

 

本文使用大小为n的数组d记录每一层的结果，达到了O（n）的空间复杂度要求。

 

 

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