规范格式

leetcode/0120.Triangle/120. Triangle.go

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+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]
+}
+
+func min(a int, b int) int {
+	if a > b {
+		return b
+	}
+	return a
+}
+
+// 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
+}

leetcode/0120.Triangle/120. Triangle_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question120 struct {
+	para120
+	ans120
+}
+
+// para 是参数
+// one 代表第一个参数
+type para120 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans120 struct {
+	one int
+}
+
+func Test_Problem120(t *testing.T) {
+
+	qs := []question120{
+		question120{
+			para120{[][]int{[]int{2}, []int{3, 4}, []int{6, 5, 7}, []int{4, 1, 8, 3}}},
+			ans120{11},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 120------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans120, q.para120
+		fmt.Printf("【input】:%v       【output】:%v\n", p, minimumTotal(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0120.Triangle/README.md

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+# [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：