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https://github.com/youngyangyang04/leetcode-master.git
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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)
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.idea/.gitignore
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# 默认忽略的文件
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/shelf/
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/workspace.xml
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# 基于编辑器的 HTTP 客户端请求
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/httpRequests/
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# Datasource local storage ignored files
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/dataSources/
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/dataSources.local.xml
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.idea/leetcode-master.iml
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.idea/leetcode-master.iml
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<?xml version="1.0" encoding="UTF-8"?>
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<module type="WEB_MODULE" version="4">
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<component name="Go" enabled="true" />
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<component name="NewModuleRootManager">
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<content url="file://$MODULE_DIR$" />
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<orderEntry type="inheritedJdk" />
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<orderEntry type="sourceFolder" forTests="false" />
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</component>
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</module>
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.idea/modules.xml
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.idea/modules.xml
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<?xml version="1.0" encoding="UTF-8"?>
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<project version="4">
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<component name="ProjectModuleManager">
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<modules>
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<module fileurl="file://$PROJECT_DIR$/.idea/leetcode-master.iml" filepath="$PROJECT_DIR$/.idea/leetcode-master.iml" />
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</modules>
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</component>
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</project>
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<?xml version="1.0" encoding="UTF-8"?>
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<project version="4">
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<component name="VcsDirectoryMappings">
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<mapping directory="$PROJECT_DIR$" vcs="Git" />
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</component>
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</project>
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@ -284,6 +284,33 @@ func min(a, b int) int {
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return b
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return b
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}
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}
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```
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```
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``` GO
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第二种思路: dp[i]表示从i层起跳所需要支付的最小费用
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递推公式:
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i<n :dp[i] = min(dp[i-1],dp[i-2])+cost[i]
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i==n:dp[i] = min(dp[i-1],dp[i-2]) (登顶)
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func minCostClimbingStairs(cost []int) int {
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n := len(cost)
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dp := make([]int, n+1)
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dp[0], dp[1] = cost[0], cost[1]
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for i := 2; i <= n; i++ {
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if i < n {
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dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
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} else {
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dp[i] = min(dp[i-1], dp[i-2])
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}
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}
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return dp[n]
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### Javascript
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### Javascript
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```Javascript
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```Javascript
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