120. Triangle【M】【95】【leetcode】

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#leetcode #动态规划 #python

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本文介绍了一种求解给定三角形结构中从顶点到底部的最小路径和问题的高效算法。通过动态规划的方法，仅使用 O(n) 的额外空间来存储中间结果，其中 n 是三角形总行数。文章提供了详细的实现步骤及 Python 代码示例。

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.

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对于题目给的输入，每次存从顶到当前节点的最小值

保留两行即可

class Solution(object):
    def minimumTotal(self, triangle):
        t = triangle
        l = len(t)
        if l == 1 :
            return t[0][0]
        res1 = [99999] * l
        res2 = [99999] * l

        res1[0] = t[0][0]
        for i in range(1,l):
            res2[0] = res1[0] + t[i][0]
            #print res1,res2
            for j in range(1,len(t[i])):
                #print i,j,t[i][j],res1[j-1],res1[j]
                res2[j] = min(res1[j],res1[j - 1]) + t[i][j]

            #print res1,res2
            res1 = res2[:]
        return min(res2)