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Problem solution · Java

Wiggle Sort II

Wiggle Sort II: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
46 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Wiggle Sort II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 46 lines of Java from the credited upstream file 324.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeWiggle Sort II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public void wiggleSort(int[] nums) {    final int n = nums.length;    final int median = findKthLargest(nums, (n + 1) / 2);    for (int i = 0, j = 0, k = n - 1; i <= k;)      if (nums[A(i, n)] > median)        swap(nums, A(i++, n), A(j++, n));      else if (nums[A(i, n)] < median)        swap(nums, A(i, n), A(k--, n));      else        ++i;  }   private int A(int i, int n) {    return (1 + 2 * i) % (n | 1);  }   // Same as 215. Kth Largest Element in an Array  private int findKthLargest(int[] nums, int k) {    return quickSelect(nums, 0, nums.length - 1, k);  }   private int quickSelect(int[] nums, int l, int r, int k) {    final int pivot = nums[r];     int nextSwapped = l;    for (int i = l; i < r; ++i)      if (nums[i] >= pivot)        swap(nums, nextSwapped++, i);    swap(nums, nextSwapped, r);     final int count = nextSwapped - l + 1; // the number of `nums` >= pivot    if (count == k)      return nums[nextSwapped];    if (count > k)      return quickSelect(nums, l, nextSwapped - 1, k);    return quickSelect(nums, nextSwapped + 1, r, k - count);  }   private void swap(int[] nums, int i, int j) {    final int temp = nums[i];    nums[i] = nums[j];    nums[j] = temp;  }}

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