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

posted @ 2015-12-10 12:00  YRB  阅读(593)  评论(0编辑  收藏  举报