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;
}
}

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