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Lines changed: 42 additions & 0 deletions

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	`1`	`+package main`
	`2`	`+`
	`3`	`+/**`
	`4`	`+回溯算法搞定`
	`5`	`+*/`
	`6`	`+`
	`7`	`+varres [][]string`
	`8`	`+`
	`9`	`+funcpartition(sstring) [][]string {`
	`10`	`+res= [][]string{}`
	`11`	`+iflen(s)<1 {`
	`12`	`+returnres`
	`13`	`+}`
	`14`	`+helper(s,0, []string{})`
	`15`	`+returnres`
	`16`	`+}`
	`17`	`+funchelper(sstring,startint,tmp []string) {`
	`18`	`+ifstart==len(s) {`
	`19`	`+m:=make([]string,len(tmp))`
	`20`	`+copy(m,tmp)`
	`21`	`+res=append(res,m)`
	`22`	`+return`
	`23`	`+}`
	`24`	`+fori:=start;i<len(s);i++ {`
	`25`	`+ifisPalindrome(s,start,i) {`
	`26`	`+tmp=append(tmp,s[start:i+1])`
	`27`	`+helper(s,i+1,tmp)`
	`28`	`+tmp=tmp[:len(tmp)-1]`
	`29`	`+}`
	`30`	`+}`
	`31`	`+}`
	`32`	`+`
	`33`	`+funcisPalindrome(sstring,start,endint)bool {`
	`34`	`+forstart<end {`
	`35`	`+ifs[start]!=s[end] {`
	`36`	`+returnfalse`
	`37`	`+}`
	`38`	`+start++`
	`39`	`+end--`
	`40`	`+}`
	`41`	`+returntrue`
	`42`	`+}`

Lines changed: 29 additions & 0 deletions

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`@@ -0,0 +1,29 @@`
	`1`	`+package main`
	`2`	`+`
	`3`	`+typeListNodestruct {`
	`4`	`+Valint`
	`5`	`+Next*ListNode`
	`6`	`+}`
	`7`	`+`
	`8`	`+/**`
	`9`	`+用快慢指针的方式，快指针每次走两步，慢指针每次走一步，相遇的时候，就是环的位置`
	`10`	`+此时slow停的位置到环入口的距离，等于头结点到环入口的距离（https://blog.csdn.net/jhlovetll/article/details/85255708）`
	`11`	`+所以两个结点前进相同的步数，head就位于环的入口处了`
	`12`	`+*/`
	`13`	`+funcdetectCycle(headListNode)ListNode {`
	`14`	`+ifhead==nil\|\|head.Next==nil {`
	`15`	`+returnnil`
	`16`	`+}`
	`17`	`+slow,fast:=head.Next,head.Next.Next`
	`18`	`+forfast!=nil&&fast.Next!=nil&&slow!=fast {`
	`19`	`+slow,fast=slow.Next,fast.Next.Next`
	`20`	`+}`
	`21`	`+//没有环`
	`22`	`+ifslow!=fast {`
	`23`	`+returnnil`
	`24`	`+}`
	`25`	`+forslow!=head {`
	`26`	`+slow,head=slow.Next,head.Next`
	`27`	`+}`
	`28`	`+returnslow`
	`29`	`+}`

Lines changed: 47 additions & 0 deletions

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	`1`	`+package main`
	`2`	`+`
	`3`	`+typeTreeNodestruct {`
	`4`	`+Valint`
	`5`	`+Left*TreeNode`
	`6`	`+Right*TreeNode`
	`7`	`+}`
	`8`	`+`
	`9`	`+/**`
	`10`	`+用一个slice实现stack结构，首先将root和root的左子树入栈`
	`11`	`+出栈的同时，将右子树入栈`
	`12`	`+*/`
	`13`	`+`
	`14`	`+typeBSTIteratorstruct {`
	`15`	`+stack []*TreeNode`
	`16`	`+}`
	`17`	`+`
	`18`	`+funcConstructor(root*TreeNode)BSTIterator {`
	`19`	`+stack:=make([]*TreeNode,0,128)`
	`20`	`+res:=BSTIterator{`
	`21`	`+stack:stack,`
	`22`	`+}`
	`23`	`+res.push(root)`
	`24`	`+returnres`
	`25`	`+}`
	`26`	`+`
	`27`	`+/** @return the next smallest number */`
	`28`	`+func (this*BSTIterator)Next()int {`
	`29`	`+size:=len(this.stack)`
	`30`	`+vartop*TreeNode`
	`31`	`+this.stack,top=this.stack[:size-1],this.stack[size-1]`
	`32`	`+this.push(top.Right)`
	`33`	`+returntop.Val`
	`34`	`+`
	`35`	`+}`
	`36`	`+`
	`37`	`+/** @return whether we have a next smallest number */`
	`38`	`+func (this*BSTIterator)HasNext()bool {`
	`39`	`+returnlen(this.stack)>0`
	`40`	`+}`
	`41`	`+`
	`42`	`+func (thisBSTIterator)push(rootTreeNode) {`
	`43`	`+forroot!=nil {`
	`44`	`+this.stack=append(this.stack,root)`
	`45`	`+root=root.Left`
	`46`	`+}`
	`47`	`+}`

Lines changed: 24 additions & 0 deletions

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	`1`	`+package main`
	`2`	`+`
	`3`	`+/**`
	`4`	+如果能将A，B分开到二个数组中，那显然符合“异或”解法的关键点了。因此这个题目的关键点就是将A，B分开到二个数组中。由于A，B肯定是不相等的，因此在二进制上必定有一位是不同的。根据这一位是0还是1可以将A，B分开到A组和B组。而这个数组中其它数字要么就属于A组，要么就属于B组。再对A组和B组分别执行“异或”解法就可以得到A，B了。而要判断A，B在哪一位上不相同，只要根据A异或B的结果就可以知道了，这个结果在二进制上为1的位就说明A，B在这一位上是不相同的。
	`5`	`+*/`
	`6`	`+funcsingleNumber(nums []int) []int {`
	`7`	`+res:=make([]int,2)`
	`8`	`+xor:=nums[0]`
	`9`	`+fori:=1;i<len(nums);i++ {`
	`10`	`+xor^=nums[i]`
	`11`	`+}`
	`12`	`+bit:=xor &^ (xor-1)`
	`13`	`+n1,n2:=0,0`
	`14`	`+for_,v:=rangenums {`
	`15`	`+ifv&bit>0 {`
	`16`	`+n1^=v`
	`17`	`+}else {`
	`18`	`+n2^=v`
	`19`	`+}`
	`20`	`+}`
	`21`	`+res[0]=n1`
	`22`	`+res[1]=n2`
	`23`	`+returnres`
	`24`	`+}`

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