解法一:just a joke :) classSolution {public:voidnextPermutation(vector<int> &num) { next_permutation(num.begin(), num.end()); } }; 解法二: 1、如果数组为降序,则根据题意,升序排序后返回。 2、如果数组为升序,则交换最后两个元素后返回。 3、举例来说明: [1,2,5,4,3]-->[1,3,2,4,5...
7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 classSolution { public: vector<int>v1; vector<vector<int>>v; vector<vector<int> > permuteUnique(vector<int> &num) { v.clear(); sort(num.begin(),num....
classSolution{public:vector<vector<int>>permute(vector<int>&nums){intamount=1;vector<vector<int>>res;for(inti=2;i<=(int)nums.size();i++)amount*=i;//计算不同排列的数量,为n!for(inti=0;i<amount;i++,nextPermutation(nums))res.push_back(nums);//生成n!种不同的排列方式returnres;}priv...
23. 欢迎查看我的Github(https:///gavinfish/LeetCode-Python) 来获得相关源代码。
class Solution: def getPermutation(self, n, k): """ :type n: int :type k: int :rtype: str """ nums = [str(i) for i in range(1, n+1)] res = [] denominator, numerator = k, k while: numerator = math.factorial(n-1) ...
https://leetcode.com/problems/next-permutation/ 参考:http://fisherlei.blogspot.hk/2012/12/leetcode-next-permutation.html 算法巧妙。记住那个算法流程图 思路: 总体来说就是从后往前找到第一个递减的数,然后在这个数的右半部分中找到一个比这个数大的最小值,交换这两个值,然后对右半部分排序就可以了.这...
Can you solve this real interview question? Letter Case Permutation - Given a string s, you can transform every letter individually to be lowercase or uppercase to create another string. Return a list of all possible strings we could create. Return the
classSolution{public:vector<vector<int>>solution(vector<int>&nums){vector<vector<int>>res;vector<int>one;if(nums.size()==1)//如果数组只剩一个元素,则递归结束{one.push_back(nums[0]);res.push_back(one);returnres;}for(int i=0;i<nums.size();i++)//每次从当前nums里取第i个数{vector...
代码 测试用例 测试结果 测试结果 全部题解 题解不存在 请查看其他题解 C++ 智能模式 1 2 3 4 5 6 class Solution { public: vector<string> letterCasePermutation(string s) { } }; 已存储 行1,列 1 Case 1Case 2 s = "a1b2" 1 2 "a1b2" "3z4" Source ...
http://bangbingsyb.blogspot.com/2014/11/leetcode-next-permutation.html 讲的比我清楚。 ** 总结:Array, 在草稿纸上多画画例子,就会有思路了。 ** Anyway, Good luck, Richardo! My code: publicclassSolution{publicvoidnextPermutation(int[]nums){if(nums==null||nums.length<=1)return;intchange=-1...