Leetcode "Triangle" Python implementation

title :

Given a triangle, find the minimum path sum from top to bottom. Each step of the 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 11 is (i.e., 2 + 3 + 5 + 1 = 11).

Note:
Bonus Point If you be able to does this using only O(n) extra space, where n is the total Number of rows in the triangle.

code : OJ Test via runtime:82 ms

1 classSolution:2     #@param triangle, a list of lists of integers3     #@return An integer4     defminimumtotal (self, triangle):5         #Special Case6         ifLen (triangle) = =0:7             return08         #DP Visit9Level =len (triangle)TenDP = [0 forIinchrange (level)] One          forIinchRange (level): A              forJinchRange (len (triangle[i]) -1,-1,-1): -                 ifJ==len (Triangle[i])-1 : -DP[J] = dp[j-1] +Triangle[i][j] the                 elifj==0: -Dp[0] = dp[0] +Triangle[i][0] -                 Else: -Dp[j] = min (dp[j-1],dp[j]) +Triangle[i][j] +         returnMIN (DP)

Ideas :

Typical dynamic programming, like the unique path, see details

Http://www.cnblogs.com/xbf9xbf/p/4250359.html

In addition, this problem has a bonus, how to use as little as possible extra space.

The general DP idea is to define an O (n) space, with additional space such as a triangle size.

This is only used for the size of the bottom layer of the triangle.

When traversing the layer I, the space storage of Dp[1:len (Triangle[i]) is used to store the minimum and the nodes to the first level of each node.

Note here: You can save the array space by traversing backwards. This is a common operation technique for array, see details

Http://www.cnblogs.com/xbf9xbf/p/4240257.html

Continuous brush problem, can be more related to the skills of the front and back, the ability to improve the code has a certain help.

Leetcode "Triangle" Python implementation