[GeeksForGeeks] Sort an array in wave form

最新推荐文章于 2025-12-10 23:46:53 发布

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原文链接：http://www.cnblogs.com/lz87/p/7208992.html

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#runtime #java #开发工具

本文介绍两种方法来将未排序的整数数组转换为波浪形数组，方法一使用排序并相邻元素交换，方法二仅需确保偶数位置的元素大于等于其相邻的奇数位置元素。

Given an unsorted array of integers, sort the array into a wave like array. An array arr[0...n-1] is sorted in wave form

if arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= ....

Solution 1. O(n * logn) runtime, using sorting

1. sort the input array.

2. swap all adjacent elements.

 1 import java.util.Arrays;
 2 public class Solution {
 3     public void sortInWaveForm(int[] nums){
 4         if(nums == null){
 5             return;
 6         }
 7         Arrays.sort(nums);
 8         for(int i = 0; i < nums.length - 1; i += 2){
 9             swap(nums, i, i + 1);
10         }
11     }
12     private void swap(int[] nums, int idx1, int idx2){
13         int temp = nums[idx1];
14         nums[idx1] = nums[idx2];
15         nums[idx2] = temp;
16     }
17     public static void main(String[] args){
18         int[] nums1 = {3, 4, 2, 1, 5};
19         int[] nums2 = {3, 5, 2, 4, 1, 6};
20         Solution sol = new Solution();
21         sol.sortInWaveForm(nums1);
22         for(int i = 0; i < nums1.length; i++){
23             System.out.print(nums1[i] + " ");
24         }
25         System.out.println();
26         sol.sortInWaveForm(nums2);
27         for(int i = 0; i < nums2.length; i++){
28             System.out.print(nums2[i] + " ");
29         }    
30     }
31 }

Solution 2. O(n) runtime

Idea: we only need to make sure that all even positioned elements are >= than their adjacent odd elements. No need to worry about odd positioned elements.

Algorithm:

Traverse all even positioned elements of input array, and do the following.

a. If current element is smaller than previous odd positioned element, swap previous and current.

b. If current element is smaller than next odd positioned element, swap next and current.

 1 public class Solution {
 2     public void sortInWaveForm(int[] nums){
 3         if(nums == null){
 4             return;
 5         }
 6         for(int i = 0; i < nums.length; i += 2){
 7             if(i > 0 && nums[i] < nums[i - 1]){
 8                 swap(nums, i, i - 1);
 9             }
10             if(i < nums.length - 1 && nums[i] < nums[i + 1]){
11                 swap(nums, i, i + 1);
12             }
13         }
14     }
15     private void swap(int[] nums, int idx1, int idx2){
16         int temp = nums[idx1];
17         nums[idx1] = nums[idx2];
18         nums[idx2] = temp;
19     }
20 }