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.
分析：该题可以用暴力搜索的方式将所有path的和存在一个vector中，但这样的空间复杂度是O(2^n)。为了使空间复杂度降到O(n)，我们可以用动态规划里record的方式，这里的record我们只需记录到上一行每个位置path的最小sum，然后通过这个计算到本行每个位置path的最小sum。代码如下：
1 class Solution { 2 public: 3 int minimumTotal(vector<vector<int> > &triangle) { 4 vector<int> pre_sum, cur_sum; 5 pre_sum.push_back(triangle[0][0]); 6 cur_sum.push_back(triangle[0][0]); 7 for(int i = 1; i < triangle.size(); i++){ 8 cur_sum.clear(); 9 for(int j = 0; j < triangle[i].size(); j++){ 10 if(j == 0) cur_sum.push_back(pre_sum[0]+triangle[i][j]); 11 else if(j == triangle[i].size()-1) cur_sum.push_back(pre_sum.back()+triangle[i][j]); 12 else cur_sum.push_back(min(pre_sum[j-1]+triangle[i][j],pre_sum[j]+triangle[i][j])); 13 } 14 pre_sum = cur_sum; 15 } 16 return *min_element(cur_sum.begin(),cur_sum.end()); 17 } 18 };
原文：http://www.cnblogs.com/Kai-Xing/p/3915845.html