题目：
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.
思路：
虽然说题目说的是要自顶向下，但是其实自底向上的效果是一致的。状态转移是 triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1])（这里不开辟新的空间，如果没有影响的时候可以重用triangle）
代码：C++
class Solution { public: int minimumTotal(vector<vector<int>>& triangle) { for(int i = triangle.size()-2; i >= 0; i--){ for(int j = 0; j <= i; j++){ triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1]); } } return triangle[0][0]; } };改进：参考leetcode的discuss
自顶向下：
class Solution { public: int minimumTotal(vector<vector<int>>& triangle) { int n = triangle.size(); vector<int>dp(n + 1); for(int i = n - 1; i >= 0; i--) for(int j = 0; j <= i; j++) dp[j] = triangle[i][j] + min(dp[j], dp[j + 1]); return dp[0]; } };自底向上:
class Solution { public: int minimumTotal(vector<vector<int>>& triangle) { int n = triangle.size(); vector<int>dp(n); for(int i = 0; i < n; i++){ if(i != 0) dp[i] = dp[i - 1] + triangle[i][i]; for(int j = i - 1; j > 0; j--) dp[j] = triangle[i][j] + min(dp[j], dp[j - 1]); dp[0] = dp[0] + triangle[i][0]; } return *min_element(dp.begin(), dp.end()); } };