• 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

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/5035540.html
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