Can you solve this real interview question? Next Permutation - A permutation of an array of integers is an arrangement of its members into a sequence or linear order. * For example, for arr = [1,2,3], the following are all the permutations of arr: [1,2
代码: classSolution {public:voidnextPermutation(vector<int>&nums) {intlen=nums.size();intpos=-1;for(inti=len-1; i>0; i--) {if(nums[i]>nums[i-1]) { pos=i-1;break; } }if(pos==-1) { reverse(nums.begin(),nums.end());return; }for(inti=len-1; i>pos;i--) {if(nums[...
// 16:20 class Solution { public: void nextPermutation(vector<int> &num) { int n = num.size(); if (n <= 1) return; int q = n - 2; while (q >= 0 && num[q] >= num[q+1]) q--; if (q < 0) { reverse(num.begin(), num.end()); return; } int i; for (i=n-...
class Solution(object): def nextPermutation(self, nums): """ :type nums: List[int] :rtype: void Do not return anything, modify nums in-place instead. """ length = len(nums) targetIndex = 0 changeIndex = 0 for i in range(length - 1, 0, -1): if nums[i] > nums[i - 1]:...
[LeetCode] next_permutation 概念 全排列的生成算法有很多种,有递归遍例,也有循环移位法等等。C++/STL中定义的next_permutation和prev_permutation函数则是非常灵活且高效的一种方法,它被广泛的应用于为指定序列生成不同的排列。本文将详细的介绍prev_permutation函数的内部算法。
然后最后想说的是,LeetCode的Discuss区里面有人用了四种解决方案解决这个问题,最让我印象深刻的是,其中一种方案只用了一条语句,因为C++的algorithm库里面有现成的next_permutation(First, last)方法解决这个问题。
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=-...
文章作者:Tyan博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution Reference https://leetcode.com/problems/next-permutation/description/
easily using this function. Hints: sortnumsin ascending order, add it to the result of all permutations and then repeatedly generate the next permutation and add it ... until we get back to the original sorted condition. If you want to learn more, please visitthis solutionandthat solution....
public class Solution { public void nextPermutation(int[] nums) { if(nums.length <= 1){ return; } int i = nums.length - 2; // 找到第一个下降点,我们要把这个下降点的值增加一点点 // 对于511这种情况,要把前面两个1都跳过,所以要包含等于 ...