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| 1 | +package com.algorithm.study.demo.algorithm; |
| 2 | + |
| 3 | +import java.util.Arrays; |
| 4 | + |
| 5 | +/** |
| 6 | + * 给定一个正整数，实现一个方法来求出离该整数最近的大于自身的“换位数”。 |
| 7 | + */ |
| 8 | +public class DictProject { |
| 9 | + public static void main(String[] args) { |
| 10 | + int[] result=findNearestNumber(new int[]{1,2,3,5,4}); |
| 11 | + for (int i=0;i<result.length;i++){ |
| 12 | + System.out.println(result[i]); |
| 13 | + } |
| 14 | + } |
| 15 | + //主流程，返回最近一个大于自身的相同数字组成的整数。 |
| 16 | + public static int [] findNearestNumber( int [] numbers){ |
| 17 | + |
| 18 | +//拷贝入参，避免直接修改入参 |
| 19 | + |
| 20 | + int [] numbersCopy = Arrays.copyOf(numbers, numbers.length); |
| 21 | +//1.从后向前查看逆序区域，找到逆序区域的前一位，也就是数字置换的边界 |
| 22 | + int index = findTransferPoint(numbersCopy); |
| 23 | + |
| 24 | +//如果数字置换边界是0，说明整个数组已经逆序，无法得到更大的相同数字组成的整数，返回自身 |
| 25 | + if(index ==0){return null;} |
| 26 | +//2.把逆序区域的前一位和逆序区域中刚刚大于它的数字交换位置 |
| 27 | + exchangeHead(numbersCopy, index); |
| 28 | +//3.把原来的逆序区域转为顺序 |
| 29 | + reverse(numbersCopy, index); |
| 30 | + |
| 31 | + return numbersCopy; |
| 32 | + } |
| 33 | + private static int findTransferPoint(int[] numbers){ |
| 34 | + for (int i=numbers.length-1;i>0;i--){ |
| 35 | + if (numbers[i]>numbers[i-1]){ |
| 36 | + return i; |
| 37 | + } |
| 38 | + } |
| 39 | + return 0; |
| 40 | + } |
| 41 | + private static void exchangeHead(int[] numbers,int index){ |
| 42 | + int head=numbers[index-1]; |
| 43 | + for (int i=numbers.length-1;i>0;i--){ |
| 44 | + if (head<numbers[i]){ |
| 45 | + numbers[index-1]=numbers[i]; |
| 46 | + numbers[i]=head; |
| 47 | + break; |
| 48 | + } |
| 49 | + } |
| 50 | + } |
| 51 | + private static int[] reverse(int[] num,int index){ |
| 52 | + for (int i=index,j=num.length-1;i<j;i++,j--){ |
| 53 | + int temp=num[i]; |
| 54 | + num[i]=num[j]; |
| 55 | + num[j]=temp; |
| 56 | + } |
| 57 | + return num; |
| 58 | + } |
| 59 | +} |
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