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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.
算法:首先想到的是DFS,但是超时了代码如下:
class Solution {public: int result; int minimumTotal(vector<vector<int> > &triangle) { result=0; doit(triangle,0,0,0); return result; } void doit(vector<vector<int>> &triangle,int level,int key,int sum) { if(level==triangle.size()) { if(sum>result) result=sum; } if(level<triangle.size()-1) { doit(triangle,level+1,key,sum+triangle[level][key]); doit(triangle,level+1,key+1,sum+triangle[level][key]); } }};
在网上查到的动态规划算法,从下到上,要求第i层到最底层路径的最小和,只要知道i+1层到最底层的最小路径和,代码如下:
class Solution {public: int minimumTotal(vector<vector<int> > &triangle) { int len=triangle.size(); if(len==0) return -1; if(triangle[len-1].size()!=len) return -1; int *temp=new int[len]; for(int i=0;i<len;i++) { temp[i]=triangle[len-1][i]; } for(int i=len-2;i>=0;i--) { for(int j=0;j<=i;j++) { temp[j]=triangle[i][j]+min(temp[j],temp[j+1]); } } return temp[0]; } };
Triangle <leetcode>
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