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+1
y_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]