120. Triangle

120. Triangle

0. 參考文獻

序號	文獻
1	Leetcode-120-Triangle
2	[eetCode 120 - Triangle]

1.題目

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

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.

2. 思路

本題求解的是給到的一組三角形矩陣中，從上到下的路徑中和最小的是多少。這裏咱們反過來求解，自底向上，設dp[i]是從底部開始到頂部的第條路徑的和。這裏怎麼理解呢？假如，三角形的底部有4個元素，那麼從底部出發到達頂部的路徑應該是有4條。所以dp[i]表明了第條路徑的和。由於題目中的規則是在第i層的第j個數字，往下只能走i+1,j 或者i+1,j+1 2個數字。所以，dp[i] = 當前數字 + min(dp[i],dp[i+1])。到這裏思路就很清晰了，咱們來看下實現。

3. 實現

class Solution(object):
    def minimumTotal(self, triangle):
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        dp = [0] *( len(triangle)  +   1 )
        
        for i in range(len(triangle)-1,-1,-1):
            
            for j in range(0,i+1):
                
                dp[j] = triangle[i][j] + min(dp[j],dp[j+1])
            print dp
        return dp[0]