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
假设原来就是字典序排列中最大的。将其又一次排列为字典序最小的排列 不要申请额外的空间 小心数组越界问题 函数没有返回值,直接改动列表 样例: 输入: [1,2,3] 输出: [1,3,2] 输入: [3,2,1] 输出: [1,2,3] 解题思路 通过一个样例来说明,原数组为[1,7,3,4,1]。我们想要找到比173421大一点的...
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。
// 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-...
leetcode 之 Next Permutation 题目描述: Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order)....
leetcode -- Next Permutation -- 重点常考 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,...
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...
Given an array of integersnums,find the next permutation ofnums. The replacement must bein placeand use only constant extra memory. 只能在原地修改,不能使用额外的数组空间。即算法的空间复杂度O(1)。 Example 1: Input:nums = [1,2,3]
[LeetCode] Next Permutation 简介:Well, in fact the problem of next permutation has been studied long ago. From the Wikipedia page, in the 14th century, a man named Narayana Pandita gi... Well, in fact the problem of next permutation has been studied long ago. From theWikipedia page, ...
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 while(i >= 0 && nums[i] >= nums[i + 1]){ ...