動機

經典的dp

Problem

Given a triangle array, return the minimum path sum from top to bottom.

For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.

 

Example 1:

Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]Output: 11Explanation: The triangle looks like:   2  3 4 6 5 74 1 8 3The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).

Example 2:

Input: triangle = [[-10]]Output: -10

 

Constraints:

  • 1 <= triangle.length <= 200
  • triangle[0].length == 1
  • triangle[i].length == triangle[i - 1].length + 1
  • -104 <= triangle[i][j] <= 104

 

Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?

Sol

每個點都是由上面的兩個點的總和構成，遞迴 每個點都是index，dp

def legal(i,len)
   i >= 0 && i < len 
end
def minimum_total(tri)
    i = 1
    while i < tri.size
        tri[i].each_with_index do |n,j|
            maxN = Float::INFINITY
            maxN = [maxN, tri[i-1][j]].min   if legal(j  ,tri[i-1].size)
            maxN = [maxN, tri[i-1][j-1]].min if legal(j-1,tri[i-1].size)
            tri[i][j] += maxN
        end
        #p tri[i]
        i += 1
    end
    tri[-1].min
end