更新0746.使用最小花费爬楼梯.md 提供Go版本解法的新思路(dp[i]表示从i层起跳所需要支付的最小费用)

.idea/.gitignore

@@ -0,0 +1,8 @@
+# 默认忽略的文件
+/shelf/
+/workspace.xml
+# 基于编辑器的 HTTP 客户端请求
+/httpRequests/
+# Datasource local storage ignored files
+/dataSources/
+/dataSources.local.xml

.idea/leetcode-master.iml

@@ -0,0 +1,9 @@
+<?xml version="1.0" encoding="UTF-8"?>
+<module type="WEB_MODULE" version="4">
+  <component name="Go" enabled="true" />
+  <component name="NewModuleRootManager">
+    <content url="file://$MODULE_DIR$" />
+    <orderEntry type="inheritedJdk" />
+    <orderEntry type="sourceFolder" forTests="false" />
+  </component>
+</module>

.idea/modules.xml

@@ -0,0 +1,8 @@
+<?xml version="1.0" encoding="UTF-8"?>
+<project version="4">
+  <component name="ProjectModuleManager">
+    <modules>
+      <module fileurl="file://$PROJECT_DIR$/.idea/leetcode-master.iml" filepath="$PROJECT_DIR$/.idea/leetcode-master.iml" />
+    </modules>
+  </component>
+</project>

.idea/vcs.xml

@@ -0,0 +1,6 @@
+<?xml version="1.0" encoding="UTF-8"?>
+<project version="4">
+  <component name="VcsDirectoryMappings">
+    <mapping directory="$PROJECT_DIR$" vcs="Git" />
+  </component>
+</project>

problems/0746.使用最小花费爬楼梯.md

@@ -284,6 +284,33 @@ func min(a, b int) int {
     return b
 }
 ```
+``` GO
+第二种思路: dp[i]表示从i层起跳所需要支付的最小费用
+递推公式:
+i<n :dp[i] = min(dp[i-1],dp[i-2])+cost[i]
+i==n:dp[i] = min(dp[i-1],dp[i-2]) (登顶)
+
+func minCostClimbingStairs(cost []int) int {
+	n := len(cost)
+	dp := make([]int, n+1)
+	dp[0], dp[1] = cost[0], cost[1]
+	for i := 2; i <= n; i++ {
+		if i < n {
+			dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
+		} else {
+			dp[i] = min(dp[i-1], dp[i-2])
+		}
+	}
+	return dp[n]
+}
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+```
+
 
 ### Javascript 
 ```Javascript