#507 Wiggle Sort II

题目描述:

Given an unsorted array nums, reorder it such that

nums[0] < nums[1] > nums[2] < nums[3]....

Notice

You may assume all input has valid answer.

Have you met this question in a real interview?

Yes

Example

Given nums = [1, 5, 1, 1, 6, 4], one possible answer is[1, 4, 1, 5, 1, 6].

Given nums = [1, 3, 2, 2, 3, 1], one possible answer is[2, 3, 1, 3, 1, 2].

题目思路：

最简单的方法是将数组排序后，按照wiggle sort的规律重新分配数字。但是题目的要求为O(n)，而quick sort或merge sort的时间复杂度均为O(nlogn)，在此并不可行。这题在算法上参考了九章的java version答案：用quicksort的思想得到中间数mid，然后根据数组长度的奇偶分配element。这里要注意的是，为了避免出现相邻数字相等的情况，分配时需要将重复的mid分配在数组两头。

Mycode（AC = 83ms）：

class Solution {
public:/*** @param nums a list of integer* @return void*/  void wiggleSort(vector<int>& nums) {// Write your code hereif (nums.size() <= 1) return;vector<int> tmp(nums);int mid = partition(tmp, 0, tmp.size() - 1, tmp.size() / 2);vector<int> ans(nums.size(), mid);// assign the numbers if nums.size() is oddif (nums.size() % 2 == 1) {int l = 0, r = nums.size() - 2;for (int i = 0; i < nums.size(); i++) {if (nums[i] > mid) {ans[r] = nums[i];r -= 2;}else if (nums[i] < mid) {ans[l] = nums[i];l += 2;}}}// assign the numbers if nums.size() is evenelse {int l = 1, r = nums.size() - 2;for (int i = 0; i < nums.size(); i++) {if (nums[i] < mid) {ans[r] = nums[i];r -= 2;}else if (nums[i] > mid) {ans[l] = nums[i];l += 2;}}}nums = ans;}// use partition function to get the mid number, note that the // nums will be changed by doing thisint partition(vector<int>& nums, int start, int end, int rank) {int left = start, right = end;int now = nums[start];while (left < right) {while (left < right && nums[right] >= now) {right--;}nums[left] = nums[right];while (left < right && nums[left] <= now) {left++;}nums[right] = nums[left];}if (left - start == rank) { // if left is in the middlereturn now;}else if (left - start < rank) { // if 'left' is in left of the middlepartition(nums, left + 1, end, rank - (left - start + 1));}else {// if 'left' is in right of the middlepartition(nums, start, left - 1, rank);}}};

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#507 Wiggle Sort II

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