zoukankan      html  css  js  c++  java
  • [LeetCode#280] Wiggle Sort

    Problem:

    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].

    Analysis:

    Once you encounter a new problem, you should not feel anxious with the new setting. 
    Try to conclude a principle underlying it, you could find out it is extraordinarily easy.
    ---------------------------------------------------------
    nums[0] <= nums[1] >= nums[2] <= nums[3]....
    ---------------------------------------------------------
    For the wiggle sort, we actually have no requirments over an element's global order. 
    for an element with odd index, we just need to meet following order principle:
    nums[i-1] <= nums[i] >= nums[i+1]
    Note: i is an odd number. 
    
    Solution 1:
    public class Solution {
        public void wiggleSort(int[] nums) {
            Arrays.sort(nums);
            int len = nums.length;
            if (len <= 2)
                return;
            for (int i = 1; i < len - 1; i = i+2) {
                int temp = nums[i];
                nums[i] = nums[i+1];
                nums[i+1] = temp;
            }
            return;
        }
    }
    
    The above solution is easy. We first guarantee the elements' order befor and after nums[i]. 
    But it needs to sort the nums array first, which we could totaly abandon.
    Since we only care about an odd(indexed) element's realtive order with it neighboring elements, we all just do it all the way. 
    for (int i = 1; i < nums.length; i++) {
        if (i % 2 == 1) {
            if (nums[i-1] > nums[i])
                swap(nums, i, i-1);
        } else{
            if (nums[i] > nums[i-1])
                swap(nums, i, i-1);
        }
    }
    
    The invariant:
    we guaratee this is no violation of wiggleSort when we reach i.
    Note the start of the for loop. 
    for (int i = 1; i < nums.length; i++)

    Solution:

    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-1] > nums[i])
                        swap(nums, i, i-1);
                } else{
                    if (nums[i] > nums[i-1])
                        swap(nums, i, i-1);
                }
            }
        }
        
        
        private void swap(int[] nums, int i, int j) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }
  • 相关阅读:
    Java Math 取整的方式
    Java final 关键词修饰类、方法、变量
    Android Activity 开发常用技巧整理
    Java override 和 overload 的区别
    Android TextView 常用技巧
    Android Adb 常用命令
    Android 实现应用升级方案(暨第三方自动升级服务无法使用后的解决方案)
    Git 常用命令
    Android 常见Crash Log汇总
    Java Annotation 总结
  • 原文地址:https://www.cnblogs.com/airwindow/p/4823159.html
Copyright © 2011-2022 走看看