zoukankan      html  css  js  c++  java
  • LeetCode

    题目:

    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.
    1,2,31,3,2
    3,2,11,2,3
    1,1,51,5,1

    思路:

    摘自Stack Overflow:

    Find the highest index i such that s{i} < s{i+1}. If no such index exists, the permutation is the last permutation. Find the highest index j > i such that s{j} > s{i}. Such a j must exist, since i+1 is such an index. Swap s{i} with s{j}. Reverse all the order of all of the elements* after index i.

    package string;
    
    public class NextPermutation {
    
        public void nextPermutation(int[] nums) {
            int len;
            if (nums == null || (len = nums.length) < 2) return;
            int index1 = -1;
            for (int i = 0; i < len - 1; ++i) {
                if (nums[i + 1] > nums[i])
                    index1 = i;
            }
            
            if (index1 != -1) {
                int index2 = index1 + 1;
                for (int j = index2 + 1; j < len; ++j) {
                    if (nums[j] > nums[index1])
                        index2 = j;
                }
                
                swap(nums, index1, index2);
                reverse(nums, index1 + 1, len);
            } else {
                reverse(nums, 0, len);
            }
        }
        
        private void swap(int[] nums, int i, int j) {
            int tmp = nums[i];
            nums[i] = nums[j];
            nums[j] = tmp;
        }
        
        private void reverse(int[] nums, int start, int end) {
            for (int i = start; i < (end - start) / 2 + start; ++i) {
                swap(nums, i, end - i + start - 1);
            }
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            NextPermutation n = new NextPermutation();
            int[] nums = { 1,3,2 };
            n.nextPermutation(nums);
            for (int i = 0; i < nums.length; ++i)
                System.out.println(nums[i]);
        }
    
    }
  • 相关阅读:
    java语言yaml序列化到文件存在类型tag
    springboot项目banner生成
    mysql自动生成大量数据
    Mysql架构
    python爬虫-UnicodeDecodeError: 'utf-8' codec can't decode byte 0x8b in position 1: invalid start byte
    六 领域驱动设计-领域对象的生命周期
    五 领域驱动设计-软件中所表示的模型
    四 领域驱动设计-分离领域
    模型驱动设计的构造块
    三 领域驱动设计-运用领域模型-绑定模型和实现
  • 原文地址:https://www.cnblogs.com/null00/p/5066680.html
Copyright © 2011-2022 走看看