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89 lines
2.0 KiB
Markdown
Executable File
89 lines
2.0 KiB
Markdown
Executable File
# [120. Triangle](https://leetcode.com/problems/triangle/)
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## 题目
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Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
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For example, given the following triangle
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[
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[2],
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[3,4],
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[6,5,7],
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[4,1,8,3]
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]
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The minimum path sum from top to bottom is `11` (i.e., **2** + **3** + **5** + **1** = 11).
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**Note**:
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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.
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## 题目大意
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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。
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## 解题思路
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- 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。
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- 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：
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## 代码
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```go
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package leetcode
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import (
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"math"
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)
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// 解法一 倒序 DP，无辅助空间
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func minimumTotal(triangle [][]int) int {
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if triangle == nil {
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return 0
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}
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for row := len(triangle) - 2; row >= 0; row-- {
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for col := 0; col < len(triangle[row]); col++ {
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triangle[row][col] += min(triangle[row+1][col], triangle[row+1][col+1])
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}
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}
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return triangle[0][0]
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}
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// 解法二 正常 DP，空间复杂度 O(n)
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func minimumTotal1(triangle [][]int) int {
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if len(triangle) == 0 {
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return 0
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}
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dp, minNum, index := make([]int, len(triangle[len(triangle)-1])), math.MaxInt64, 0
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for ; index < len(triangle[0]); index++ {
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dp[index] = triangle[0][index]
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}
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for i := 1; i < len(triangle); i++ {
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for j := len(triangle[i]) - 1; j >= 0; j-- {
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if j == 0 {
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// 最左边
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dp[j] += triangle[i][0]
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} else if j == len(triangle[i])-1 {
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// 最右边
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dp[j] += dp[j-1] + triangle[i][j]
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} else {
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// 中间
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dp[j] = min(dp[j-1]+triangle[i][j], dp[j]+triangle[i][j])
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}
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}
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}
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for i := 0; i < len(dp); i++ {
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if dp[i] < minNum {
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minNum = dp[i]
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}
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}
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return minNum
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}
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``` |