題目：

Given an unsorted array?nums, reorder it?in-place?such that?nums[0] <= nums[1] >= nums[2] <= nums[3]....

For example, given?nums = [3, 5, 2, 1, 6, 4], one possible answer is?[1, 6, 2, 5, 3, 4].

鏈接：?http://leetcode.com/problems/wiggle-sort/

題解：

Wiggle排序數組。按照題意寫就可以了。 還可以簡化，要多學習Stefan的代碼。

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {public void wiggleSort(int[] nums) {for(int i = 1; i < nums.length; i++) {if(i % 2 == 1) {if(nums[i] < nums[i - 1]) {swap(nums, i);}} else {if(i != 0 && nums[i] > nums[i - 1]) {swap(nums, i);}}}}private void swap(int[] nums, int i) {int tmp = nums[i - 1];nums[i - 1] = nums[i];nums[i] = tmp;} }

?

二刷:

方法和一刷一樣。在i % 2 == 1或者 == 0的時候作交換的判斷，交換完畢以后仍然保持這是一個本地化操作就可以了。交換的時候是用 i和 i - 1來交換

Java:

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {public void wiggleSort(int[] nums) {if (nums == null || nums.length == 0) {return;}for (int i = 1; i < nums.length; i++) {if (i % 2 == 1) {if (nums[i] < nums[i - 1]) {swap(nums, i);} } else {if (nums[i] > nums[i - 1]) {swap(nums, i);}}}}private void swap(int[] nums, int i) {int tmp = nums[i - 1];nums[i - 1] = nums[i];nums[i] = tmp;} }

?

三刷:

跟前面一樣，也是直接編寫。

Java:

public class Solution {public void wiggleSort(int[] nums) {if (nums == null || nums.length < 2) return;for (int i = 1; i < nums.length; i++) {if ((i % 2 == 0 && nums[i] > nums[i - 1]) || (i % 2 == 1 && nums[i] < nums[i - 1])) {int tmp = nums[i];nums[i] = nums[i - 1];nums[i - 1] = tmp;} }} }

?

?

Reference:

https://leetcode.com/discuss/57113/java-o-n-solution

https://leetcode.com/discuss/57206/java-o-n-10-lines-consice-solution

https://leetcode.com/discuss/60824/java-python-o-n-time-o-1-space-solution-3-lines

https://leetcode.com/discuss/57118/easy-code-of-python

轉載于:https://www.cnblogs.com/yrbbest/p/5035540.html

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