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.

示例：

[[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.

题目解析：

1.此题是等腰三角形，上下之间的关系简化为上下相邻的三个数，index相邻，大小关系是在下方二选一+上方的数值，必然正确。 根据此思路，可以top-bottom,或者bottom-top，由于可以简化，所以动态规划方法。
2. 采用bottom-top方法，最后简化为1个值，所以左侧值放置两值中的小值。

代码：

普通代码，较慢：
class Solution_:def minimumTotal(self, triangle):""":type triangle: List[List[int]]:rtype: int"""all_paths=[]cur_path=[triangle[0][0]]# cur_path=[]cur_index=[0,0]self.bfs(all_paths,cur_path,cur_index,triangle)print(all_paths)sums=[sum(elem) for elem in all_paths]return min(sums)def bfs(self,all_paths,cur_path,cur_index,triangle):x_cur=cur_index[0]# cur_row=triangle[x_cur]x_threshold=len(triangle)y_cur=cur_index[1]x_next=x_cur+1y_next=[]if x_next<x_threshold:next_row = triangle[x_cur + 1]y_threshold = len(next_row)# y_next.append(y_cur)# if y_cur-1>=0:y_next.append(y_cur)if y_cur+1<y_threshold:y_next.append(y_cur+1)for elem in y_next:cur_path_pre=cur_path+[triangle[x_next][elem]]self.bfs(all_paths,cur_path_pre,[x_next,elem],triangle)else:all_paths.append(cur_path)
动态规划，简练：
class Solution(object):def minimumTotal(self, triangle):dp=triangle[-1][:]# print(dp)for i in  range(len(triangle)-2,-1,-1):for j in range(i+1):# print(i,j)dp[j]=min(dp[j],dp[j+1])+triangle[i][j]# dp=dp[:i+1]return dp[0]