Commit 4a3de00

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lixiang.2533

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modify

Change-Id: I3a0390b95cb88b7655dfee3c547f6238bbafd670

1 parent 3b8f934 commit 4a3de00Copy full SHA for 4a3de00

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Original file line number	Diff line number	Diff line change
`@@ -44,7 +44,7 @@ class Solution:`
`44`	`44`
`45`	`45`	`Tips`
`46`	`46`
`47`		`-1. 初始化这里有一点需要注意的,就是初始长度是len(num)+1,这里是针对只有两个元素的nums,如果只初始len(nums)就需要对dp[1]进行特殊处理了。dp[0]=0,dp[1]=nums[0]`
	`47`	`+1. 初始化这里有一点需要注意的,就是初始化只能初始化1个位置,如果直接用nums初始化两个位置,会存在问题,这时第二个位置就不满足最优的rob方案了。所以dp长度是l+1, 初始化一个位置`
`48`	`48`	`2. 每一步,这里因为加入了dp[0],所以有一步错位`
`49`	`49`
`50`	`50`	`$$`

Lines changed: 2 additions & 2 deletions

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`@@ -32,5 +32,5 @@`
`32`	`32`	`- 460 LFU:`
`33`	`33`	`- 707 设计链表`
`34`	`34`	`- 技巧题`
`35`		`- - 160 相交链表:可以用常规接发, 找到长度,拼接成相同长度,然后遍历判断是否有相交节点`
	`35`	`+ - 160 相交链表:`
`36`	`36`	`- 382 链表随机节点:蓄水池算法,第K个节点有1/K的概率overwrite之前sample的元素`

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