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[LeetCode][Java] Permutations II
题目: Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, [1,1,2] have the following unique permutations: [1,1,2], [1,2,1], and...
分类:编程语言   时间:2015-07-14 13:39:54    阅读次数:180
leetCode 47.Permutations II (排列组合II) 解题思路和方法
Permutations II  Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, [1,1,2] have the following unique permutations: [1,1,2], [1,...
分类:其他好文   时间:2015-07-13 14:06:17    阅读次数:130
[leedcode 60] Permutation Sequence
The set[1,2,3,…,n]contains a total ofn! unique permutations.By listing and labeling all of the permutations in order,We get the following sequence (ie...
分类:其他好文   时间:2015-07-13 13:37:19    阅读次数:127
[leedcode 47] Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations.For example,[1,1,2]have the following unique perm...
分类:其他好文   时间:2015-07-11 13:22:30    阅读次数:122
[leedcode 46] Permutations
Given a collection of numbers, return all possible permutations.For example,[1,2,3]have the following permutations:[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,...
分类:其他好文   时间:2015-07-11 12:07:21    阅读次数:117
leetcoder 46-Permutations and 47-Permutations II
Permutations Given a collection of numbers, return all possible permutations. For example, [1,2,3] have the following permutations: [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]...
分类:其他好文   时间:2015-07-09 22:45:59    阅读次数:154
047 Permutations 2
这道题与 046 Permutations 基本一样, 唯一要注意的是需要去除重复的。 方法是在每一层的选数过程中如果相邻(当然是先排序了的)的数字相同,则在这一层中不再选举。class Solution: # @param {integer[]} nums # @return {int...
分类:其他好文   时间:2015-07-09 06:18:40    阅读次数:121
LeetCode60:Permutation Sequence
The set [1,2,3,…,n] contains a total of n! unique permutations.By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3):1 “123” 2 “132” 3 “213”...
分类:其他好文   时间:2015-07-08 10:57:54    阅读次数:512
cf 251 B Playing with Permutations 暴力 分类讨论
题意: 题意是,给q数列,和s数列。然后p数列初始为1-n。然后通过p[q[i]]=p[i],或者p[i]=p[q[i]]这两种变换,问有没有可能在k次变换后刚刚p数列为s数列。并且在这k次变换过程中,p数列不能等于s数列。p数列一开始就为s数列也不行。 做法:因为两个变换是相反的,所以可以通过两次分别两种变换来抵消。计算出p通过第一种变换要多少步可以达到s数列,然后第二种变换要多少步,然后分类讨论。...
分类:其他好文   时间:2015-07-07 21:14:22    阅读次数:156
[LintCode] 全排列
递归实现:class Solution {public: /** * @param nums: A list of integers. * @return: A list of permutations. */ vector > permute(vector nu...
分类:其他好文   时间:2015-07-07 02:03:20    阅读次数:194
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