Dynamic Programming - 120. Triangle

120. Triangle

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：

题目意思是给一个二维数组，找出一条路径和是最小的，并且上一行到下一行只能走相邻的位置。很明显看出存在子问题是下一步走哪一边，使得和最小，所以使用动态规划求解，状态转移方程就是dp[i][j] = min(dp[i+1][j] + triangle[i][j], dp[i+1][j+1] + triangle[i][j]), 初始条件是最后一行没办法用方程求解，求解顺序是自顶向上，并且可以优化为一维dp，就是只用开辟最长的子数组来作为dp的数组。甚至可以直接在原数组上操作，空间复杂度优化为O(1)。

代码：

go：

/*func minimumTotal(triangle [][]int) int {

    if triangle == nil || len(triangle) == 0 {
        return 0
    }
    
    m := len(triangle)
    dp := make([][]int, m)
    for i := 0; i < m; i++ {
        dp[i] = make([]int, len(triangle[i]))
    }
    
    for i := m - 1; i >= 0; i-- {
        for j := 0; j < len(triangle[i]); j++ {
            if i == m-1 {
                dp[i][j] = triangle[i][j]
            } else {
                dp[i][j] = min(dp[i+1][j] + triangle[i][j], dp[i+1][j+1] + triangle[i][j])
            }
        }
    }
    
    return dp[0][0]
}*/

func minimumTotal(triangle [][]int) int {
    if triangle == nil || len(triangle) == 0 {
        return 0
    }
    
    m := len(triangle)
    dp := make([]int, m)
    
    for i := m - 1; i >= 0; i-- {
        for j := 0; j < len(triangle[i]); j++ {
            if i == m-1 {
                dp[j] = triangle[i][j]
            } else {
                dp[j] = min(dp[j] + triangle[i][j], dp[j+1] + triangle[i][j])
            }
        }
    }
    return dp[0]
}

func min(i, j int) int {
    if i < j {
        return i
    } else {
        return j
    }
}