31. Next Permutation (下一个全排列)

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).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

原文地址:https://www.cnblogs.com/zle1992/p/8538649.html

时间: 2024-08-29 02:40:41

31. Next Permutation (下一个全排列)的相关文章

[LeetCode]31. Next Permutation下一个排列

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). The replaceme

leetCode 31.Next Permutation (下一个字典序排序) 解题思路和方法

Next Permutation 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 orde

[leetcode]31. Next Permutation 下一个排列

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). The replaceme

LeetCode 31 Next Permutation(下一个全排列)

题目链接: https://leetcode.com/problems/next-permutation/?tab=Description Problem :寻找给定int数组的下一个全排列(要求:be in-place) 倒序查找到该数组中第一个满足后面的数字大于前面的数字的下标i (当前下标 i 指向 后面的那个比较大的数) 参考代码: package leetcode_50; /*** * * @author pengfei_zheng * 下一个全排列 */ public class S

NextPermutation,寻找下一个全排列

问题描述:给定一个数组是一个全排列,寻找下一个全排列.例如123->132, 321->123, 115->151. 算法分析:从后往前寻找顺序,找到后从往前寻找第一个大于当前元素,即第一个逆序,然后元素交换,重新sort当前元素后面的元素.如果都是逆序,reverse数组. 1 package Leecode_Permutation; 2 3 import java.util.Arrays; 4 5 public class NextPermutation { 6 public voi

Leetcode:Next Permutation 下一个排列

Next Permutation: 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 ord

【LeetCode每天一题】Next Permutation(下一个排列)

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). The replaceme

lintcode 中等题:next permutation下一个排列

题目 下一个排列 给定一个整数数组来表示排列,找出其之后的一个排列. 样例 给出排列[1,3,2,3],其下一个排列是[1,3,3,2] 给出排列[4,3,2,1],其下一个排列是[1,2,3,4] 注意 排列中可能包含重复的整数 解题 和上一题求上一个排列应该很类似 1.对这个数,先从右到左找到递增序列的前一个位置,peakInd 2.若peakInd = -1 这个数直接逆序就是答案了 3.peakInd>= 0 peakInd这个位置的所,和 peakInd 到nums.size() -1

2.1.12 Next Permutation 下一个字典序数组

对当前排列从后向前扫描,找到一对为升序的相邻元素,记为i和j(i < j).如果不存在这样一对为升序的相邻元素,则所有排列均已找到,算法结束:否则,重新对当前排列从后向前扫描,找到第一个大于i的元素k,交换i和k,然后对从j开始到结束的子序列反转,则此时得到的新排列就为下一个字典序排列.这种方式实现得到的所有排列是按字典序有序的,这也是C++ STL算法next_permutation的思想.