# [120. Triangle](https://leetcode.com/problems/triangle/)


## 题目

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.


## 题目大意

给定一个三角形，找出自顶向下的最小路径和。每一步只能移动到下一行中相邻的结点上。


## 解题思路

- 求出从三角形顶端到底端的最小和。要求最好用 O(n) 的时间复杂度。
- 这一题最优解是不用辅助空间，直接从下层往上层推。普通解法是用二维数组 DP，稍微优化的解法是一维数组 DP。解法如下：


## 代码

```go

package leetcode

import (
	"math"
)

// 解法一 倒序 DP，无辅助空间
func minimumTotal(triangle [][]int) int {
	if triangle == nil {
		return 0
	}
	for row := len(triangle) - 2; row >= 0; row-- {
		for col := 0; col < len(triangle[row]); col++ {
			triangle[row][col] += min(triangle[row+1][col], triangle[row+1][col+1])
		}
	}
	return triangle[0][0]
}

// 解法二 正常 DP，空间复杂度 O(n)
func minimumTotal1(triangle [][]int) int {
	if len(triangle) == 0 {
		return 0
	}
	dp, minNum, index := make([]int, len(triangle[len(triangle)-1])), math.MaxInt64, 0
	for ; index < len(triangle[0]); index++ {
		dp[index] = triangle[0][index]
	}
	for i := 1; i < len(triangle); i++ {
		for j := len(triangle[i]) - 1; j >= 0; j-- {
			if j == 0 {
				// 最左边
				dp[j] += triangle[i][0]
			} else if j == len(triangle[i])-1 {
				// 最右边
				dp[j] += dp[j-1] + triangle[i][j]
			} else {
				// 中间
				dp[j] = min(dp[j-1]+triangle[i][j], dp[j]+triangle[i][j])
			}
		}
	}
	for i := 0; i < len(dp); i++ {
		if dp[i] < minNum {
			minNum = dp[i]
		}
	}
	return minNum
}

```