permutations

LeetCode刷题笔录Permutations II

Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].和Permutation那题写法差不多，唯一要加的地方就是
likecool21·2014-10-15 06:00

全排列(permutations)

算法设计时候，递归技巧可以很方便地解决一些复杂的问题。求一串数字的全排列就可以用递归技巧写出程序。例如,[1,2,3]有下列排列:[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],和[3,2,1].递归过程：1，为了求[1,2,3]的全排列，可以先求子串[2,3]的全排列，最后将1加到结果中即可2，求所有子串的全排列，即去掉2求子串[1,3]，去掉3求子串[1,2]。
bdss58·2014-10-04 17:00

【线段树+字符串hash】 codeforces 213E Two Permutations

对于排列单点插入到线段树中，然后判断整个线段树区间上的hash值和匹配串的hash值是否相同。。。#include #include #include #include #include #include #include #include #include #include #include #include #include #definemaxn200005 #definemaxm30000
blankcqk·2014-09-30 11:00

Gargari and Permutations

Gargarigotboredtoplaywiththebishopsandnow,aftersolvingtheproblemaboutthem,heistryingtodomathhomework.Inamathbookhehavefoundkpermutations.Eachofthemconsistsofnumbers1, 2, ..., ninsomeorder.Nowheshouldf
猪刚烈·2014-09-24 13:00

[LeetCode][I]Permutations II

Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].publicclassSolution{ booleani
u014691362·2014-09-23 13:00

LeetCode Permutations II

题目描述：https://oj.leetcode.com/problems/permutations-ii/问题大意：给定一个带有重复数字的数组，输出其中元素的所有排列。
tonghu2010·2014-09-13 10:00

LeetCode Permutations

题目描述：https://oj.leetcode.com/problems/permutations/问题大意：给定一个数组，求出其中的所有排列。DFS可解。
tonghu2010·2014-09-13 10:00

LeetCode-Permutations

题目：https://oj.leetcode.com/problems/permutations/Givenacollectionofnumbers,returnallpossiblepermutations.Forexample
xin_jmail·2014-09-05 22:00

LeetCode——Permutations II

Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].原题链接：https://oj.leetcode.com/
ozhaohuafei·2014-09-01 13:00

UVA - 11077 Find the Permutations (置换)

Sortingisoneofthemostusedoperationsinreallife,whereComputerSciencecomesintoact.Itiswell-knownthatthelowerboundofswapbasedsortingisnlog(n).ItmeansthatthebestpossiblesortingalgorithmwilltakeatleastW(nlo
u011345136·2014-09-01 09:00

Gargari and Permutations

Gargarigotboredtoplaywiththebishopsandnow,aftersolvingtheproblemaboutthem,heistryingtodomathhomework.Inamathbookhehavefoundkpermutations.Eachofthemconsistsofnumbers1, 2, ..., ninsomeorder.Nowheshouldf
u011345136·2014-08-31 23:00

Gargari and Permutations【DP】

题目大意：给出1~n的k个排列（2pos[u][j](1 #include #include #include #defineN1010 usingnamespacestd; intpos[6][N],cnt[N],a[6][N],dp[N]; vectorq; intmain() { intn,k,ans=0; scanf("%d%d",&n,&k); for(inti=0;i
u013912596·2014-08-31 14:00

codeforce 263 div2D Gargari and Permutations

D.GargariandPermutationstimelimitpertest2secondsmemorylimitpertest256megabytesinputstandardinputoutputstandardoutputGargarigotboredtoplaywiththebishopsandnow,aftersolvingtheproblemaboutthem,heistryi
zjck1995·2014-08-30 23:00

Codeforces 463D Gargari and Permutations(BFS)

题目链接：Codeforces 463D Gargari and Permutations 题目大意：求k个序列的最长公共子序列。
阿尔萨斯·2014-08-30 21:00

LeetCode: Permutations II

思路：和一般的排列一样，不过在存入结果时，先检查一下是否已经存在这个结果了。code:classSolution{ public: voidsolvePermute(vector&num,intpos,vector>&ret){ if(pos==num.size()-1){ if(find(ret.begin(),ret.end(),num)==ret.end()) ret.push_back(n
AIvin24·2014-08-30 13:00

Permutations Permutations||

地址：https://oj.leetcode.com/problems/permutations/ https://oj.leetcode.com/problems/permutations-ii/题目简单说就是实现全排列
huruzun·2014-08-29 11:00

STL中简单算法实例sort()、next_permutation()

= "quit")     {         cout<<"Permutations of "<
xy六道·2014-08-28 19:38

Permutations

Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],and [3,2,1].首先排序1.先从右到左，找到第一个违反递减规律的数num[i]2.再从右到i+1
huangynn·2014-08-26 14:00

uva 11077 - Find the Permutations(置换)

题目链接：uva11077-FindthePermutations题目大意：给定一个1~n的排序，可以通过一系列的交换变成1，2，…,n,给定n和k，统计有多少个排列至少需要交换k次才能变成有序的序列。解题思路：给定一个序列P，可以将该序列看做是一个置换，从有序序列，开始，需要多少次回到有序序列。将P的循环分解，循环长度为1的需要0次，长度为2的需要1次，循环长度为n的需要n-1次，如果P的长度为
u011328934·2014-08-13 11:00

LeetCode——Permutations

Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],and [3,2,1].原题链接：https://oj.leetcode.com/problems/pe
ozhaohuafei·2014-08-10 09:00

Permutations II leetcode java

题目： Given a collection of numbers that might contain duplicates, return all possible unique permutations
·2014-08-08 00:00

Permutation Sequence leetcode java

unique permutations.
·2014-08-07 08:00

Permutations leetcode java

题目： Given a collection of numbers, return all possible permutations.
·2014-08-03 09:00

poj 2369 Permutations 置换水题

找循环节求lcm就够了，若答案是12345应该输出1，被坑了下。#include #include #include #include #include #include usingnamespacestd; #defineINF0x3FFFFFF #defineMAXN2222 #defineeps1e-6 inta[MAXN],p[MAXN],b[MAXN],vis[MAXN]; intgcd
t1019256391·2014-07-20 16:00

[LeetCode] Permutations II

vector>permuteUnique(vector&num){ vector>permutations; if(num.size()>permutationsWithoutLastNum=permuteUnique
HQBUPT·2014-07-12 22:00

[LeetCode] Permutation

vector>permute(vector&num){ vector>permutations; if(num.size()>permutationsWithoutLastNum=permute(num
HQBUPT·2014-07-12 21:00

Principle of Computing (Python)学习笔记(4) Combination + Yahtzee

page=permutations2代码部分http://www.codeskulptor.org/#poc_e
feliciafay·2014-07-11 06:00

[LeetCode]—Permutations II 求全排列（有重复值）

PermutationsII Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].分析：     本题在[Le
SUN20082567·2014-07-10 11:00

[LeetCode]—Permutations 求全排列

Permutations Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations
SUN20082567·2014-07-10 10:00

Permutations II

做LeetCode44.Permutations时，考虑的就是找出所有uniquepermutations，而且44中也没有考察duplicate的sample,所以同一份代码都能AC思路是num排升序
u014674776·2014-06-29 04:00

Permutations

num排升序，将其加入返回值ret中，再迭代去找它的LeetCode30.NextPermutation代码：classSolution { public: vector>permute(vector&num) { vector>ret; sort(num.begin(),num.end()); do { ret.push_back(num); }while(has_next_permutati
u014674776·2014-06-29 04:00

[leetcode] Permutations II

Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2]havethefollowinguniquepermutations:[1,1,2],[1,2,1],and[2,1,1].https://oj.leetcode.com/problems
jdflyfly·2014-06-26 10:00

[leetcode] Permutations

havethefollowingpermutations:[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],and[3,2,1].https://oj.leetcode.com/problems/permutations
jdflyfly·2014-06-26 10:00

每日算法之三十六：Permutations && Permutations II

Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],and [3,2,1].将第0个字符和从第0开始的每个字符进行交换，对于交换后的结果，再从第1个字符开始
yapian8·2014-06-25 20:00

LeetCode: Permutations

思路：经典题目，A[1,...,k]全排列的生成过程就是先将A[1]与后面的元素一一交换，与剩下的子数组所有的全排列组合起来，得到新的全排列，就比方P1,P2,...是2-k子数组的全排列，那么:A1P1A1P2....是一组，交换A1与A2，得到新的一组排列：A2P^1A2P^2...code:classSolution{ public: voidsolvePermute(vector&num,
AIvin24·2014-06-23 21:00

[leetcode]Permutation Sequence

unique permutations. By listing and labeling all of the permutatio
huntfor·2014-06-16 15:00

[leetcode]Permutation Sequence @ Python

unique permutations. By listing and labeling all of the permutations in order,
·2014-06-13 11:00

[LeetCode68]Permutations Sequence

Theset [1,2,3,…,n] containsatotalof n!uniquepermutations.Bylistingandlabelingallofthepermutationsinorder,Wegetthefollowingsequence(ie,for n =3):"123""132""213""231""312""321"Given n and k,returnthe kt
sbitswc·2014-06-13 10:00

[LeetCode47]Permutations II

Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].Analysis:similarwiththeformer
sbitswc·2014-06-13 06:00

[LeetCode69]Permutations

Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],and [3,2,1].Analysis:Recursion,useflagarraytoidentif
sbitswc·2014-06-13 06:00

[leetcode]Permutations II

新博文地址： [leetcode]PermutationsII Permutations II Given a collection of numbers that might contain duplicates
huntfor·2014-06-12 17:00

[LeetCode]Permutations II

题目描述Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].给出无重复的带有重复数字数组的全排列。解题思路在P
cjllife·2014-06-11 17:00

[LeetCode]Permutations

题目描述Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],and [3,2,1].给出数组数字的全排列解题思路采用递归。我们求整个数组的全排列，可以看成求
cjllife·2014-06-11 15:00

【leetcode】 Permutation Sequence

问题： 对于给定序列1...n，permutations共同拥有 n！个，那么随意给定k，返回第k个permutation。0 < n < 10。
·2014-06-10 15:00

[leetcode] Permutations

Permutations思路：交换+DFSclassSolution{ vector>res; intlength; public: vector>permute(vector&num){ length
lydyangliu·2014-06-03 00:00

【leetcode】 Permutation Sequence

问题：对于给定序列1...n，permutations共有n！个，那么任意给定k，返回第k个permutation。
shiquxinkong·2014-06-01 11:00

【leetcode】Permutations

问题：Givenacollectionofnumbers,returnallpossiblepermutations.Forexample,[1,2,3] havethefollowingpermutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2],and [3,2,1].分析：permutationsII 里的 一样，这里排个序就很简单，一样的
shiquxinkong·2014-05-31 23:00

【leetcode】Permutations II

问题：Givenacollectionofnumbersthatmightcontainduplicates,returnallpossibleuniquepermutations.Forexample,[1,1,2] havethefollowinguniquepermutations:[1,1,2], [1,2,1],and [2,1,1].说明：这个问题建议看下侯捷的《STL源码剖析》，这本
shiquxinkong·2014-05-31 22:00

[leetcode]Permutations II @ Python

原题地址：https://oj.leetcode.com/problems/permutations-ii/ 题意： Given a collection of numbers that
·2014-05-29 13:00

[leetcode]Permutations @ Python

原题地址：https://oj.leetcode.com/problems/permutations/ 题意： Given a collection of numbers, return all possible
·2014-05-29 12:00
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