280. Wiggle Sort

题目：

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].

链接： http://leetcode.com/problems/wiggle-sort/

题解：

Wiggle排序数组。按照题意写就可以了。还可以简化，要多学习Stefan的代码。

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:

方法和一刷一样。在i % 2 == 1或者 == 0的时候作交换的判断，交换完毕以后仍然保持这是一个本地化操作就可以了。交换的时候是用 i和 i - 1来交换

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:

跟前面一样，也是直接编写。

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

转载于:https://www.cnblogs.com/yrbbest/p/5035540.html

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