第十七周 动态规划

算法题目 ：Triangle           

算法题目描述：

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, wheren is the total number of rows in the triangle.

 

算法分析：

这道题目可以使用使用动态规划法。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。

 达到了O（n）的空间复杂度要求

算法代码（C++):

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];     }};