120. Triangle 三角形最小路径和 code

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.

思路：

1. 设置一个二维数组，最优值三角形 $dp$，并初始化数组元素为0。 $dp[i][j]$代表 了从底向上递推时，走到三角形第i行第j列的最优解。
2. 从三角形的底向上进行动态规划:
   a.动态规划边界条件：底上的最优值即为数字三角形的最后一层。
   b.利用 $i$ 循环，从倒数第二层递推至第一层，对于每层的各列，进行动态规划递推:
   第 $i$行，第j列的最优解为 $dp[i][j]$，可到达 $(i,j)$的两个位置的最优解 $dp[i+1][j]$、 $dp[i+1][j+1]$:
   $dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j]$
3. 返回 $dp[0][0]$.

$dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j]$ 可以变成一维DP：

For the i-th level:
$dp[j] = min(dp[j], dp[j+1]) + triangle[i][j]$
同样可以写作：
$dp[j] = min(dp[j]+ triangle[i][j], dp[j+1]+ triangle[i][j]) $

# 时间复杂度为O(n^2)，空间复杂度O(n)。n为三角形高度。
from typing import List
class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        n = len(triangle)
        dp = triangle[-1]
        for i in range(n-2, -1, -1): # row 倒数第二行开始遍历
            for j in range(i + 1): 
                dp[j] = min(dp[j]+triangle[i][j], dp[j+1]+triangle[i][j]) 
        return dp[0]

triangle = [
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]
solution = Solution()
solution.minimumTotal(triangle)