[Lintcode]109. Triangle/[Leetcode]120. Triangle-优快云博客

本文探讨了在给定三角形结构中寻找从顶部到底部的最小路径和问题，使用动态规划方法解决，并提供了具体代码实现。通过逆向思维，从底部开始计算，将每个节点的最小路径和更新为到其下方两节点中较小值加自身值。

109. Triangle/120. Triangle

本题难度: Medium
Topic: Dynamic Programming

Description

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

Example
example 1
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).

example 2
Given the following triangle:

[
[2],
[3,2],
[6,5,7],
[4,4,8,1]]
The minimum path sum from top to bottom is 12 (i.e., 2 + 2 + 7 + 1 = 12).

Notice
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.

我的代码

class Solution:
    """
    @param triangle: a list of lists of integers
    @return: An integer, minimum path sum
    """
    def minimumTotal(self, triangle):
        # write your code here
    
        l = len(triangle)
        if l == 0:
            return 0
        if l == 1:
            return triangle[0][0]
        res = triangle[-1]
        for i in range(l-2,0,-1):
            for j in range(len(triangle[i])):
                res[j] = triangle[i][j]+min(res[j],res[j+1])
        return triangle[0][0]+min(res[:2])

思路
从下往上，每一个都是到达下面相邻的两个元素中路径中较小的加上当前元素之和，作为到达该位置的最小路径

      [2],     ^         [2]
    [3, 4],    |       [9, 10,          10, 3]
   [6, 5, 7],  |     [7, 6, 10,         3]
  [4, 1, 8, 3] |   [4, 1,  8, 3]

注意特殊情况：
只有一个元素和空三角形