Question Definition
Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....
Example:
(1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6].
(2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3, 1, 3, 1, 2].
Note: You may assume all input has valid answer.
Follow Up: Can you do it in O(n) time and/or in-place with O(1) extra space?
Java Solution
class Solution {
public void wiggleSort(int[] nums) {
int median = findKthLargest(nums, (nums.length + 1) / 2);
int n = nums.length;
int left = 0, i = 0, right = n - 1;
while (i <= right) {
if (nums[newIndex(i,n)] > median) {
swap(nums, newIndex(left++,n), newIndex(i++,n));
}
else if (nums[newIndex(i,n)] < median) {
swap(nums, newIndex(right--,n), newIndex(i,n));
}
else {
i++;
}
}
}
private int newIndex(int index, int n) {
return (1 + 2*index) % (n | 1);
}
private int findKthLargest(int[] nums, int k) {
int start = 0;
int end = nums.length - 1;
while(true){
int partition = partition(nums, start, end);
if(partition == k - 1)
return nums[partition];
else if(partition > k - 1){
end = partition - 1;
}else{
start = partition + 1;
}
}
}
private int partition(int[] a, int lo, int hi){
int i = lo, j = hi+1; // left and right scan indices
int v = a[lo]; // partitioning item
while (true)
{ // Scan right, scan left, check for scan complete, and exchange.
while (++i < hi && a[i] > v);
while (--j > lo && v > a[j]) if (j == lo) break;
if (i >= j) break;
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
int temp = a[lo];
a[lo] = a[j];
a[j] = temp;
return j;
}
private void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
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