[leetcode] 120. Triangle

题目

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: 11
Explanation: The triangle looks like:
   2
  3 4
 6 5 7
4 1 8 3
The 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?

思路

动态规划，从低往上计算最短路径，每个节点的最短路径=该节点数值+min（左子节点的最短路径，右子节点的最短路径），计算至根节点时即可得到最短路径。

代码

python版本：

# dfs，超时
class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        def dfs(i,j):
            if i>=len(triangle) or j>=len(triangle[i]):
                return 0
            return triangle[i][j]+min(dfs(i+1,j),dfs(i+1,j+1))        
        return dfs(0,0)

# 动态规划
class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        dp = [i for i in triangle[-1]]
        for i in range(len(triangle)-2, -1, -1):
            for j in range(len(triangle[i])):
                dp[j] = triangle[i][j]+min(dp[j], dp[j+1])
        return dp[0]