Dynamic Programming - 120. Triangle
Dynamic Programming - 120. Triangle
ppxai
发布于 2020-09-23 17:27:59
发布于 2020-09-23 17:27:59
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
}
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原始发表：2019年08月01日，如有侵权请联系 cloudcommunity@tencent.com 删除
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