Wiggle Sort II(Medium) ---摇摆排序
问题: Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....
例子: (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].
方法一
排序,然后两边分别取,复杂度O(nlogn);注意排完序之后应该倒着来
public void wiggleSort(int[] nums) {
Arrays.sort(nums);
int[] temp = new int[nums.length];
int s = (nums.length + 1) >> 1;
int t = nums.length;
for (int i = 0; i < nums.length; i++) {
temp[i] = (i & 1) == 0 ? nums[--s] : nums[--t] ;
}
for (int i = 0; i < nums.length; i++)
nums[i] = temp[i];
}
方法二
用快排的思想查找中位数,然后再合并两边。
最坏复杂度O(n^2),平均复杂度O(n)
public void wiggleSort(int[] nums) {
int medium = findMedium(nums, 0, nums.length - 1, (nums.length + 1) >> 1);
int s = 0, t = nums.length - 1 , mid_index = (nums.length + 1) >> 1;
int[] temp = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
if (nums[i] < medium)
temp[s++] = nums[i];
else if (nums[i] > medium)
temp[t--] = nums[i];
}
while (s < mid_index)
temp[s++] = medium;
while (t >= mid_index)
temp[t--] = medium;
t = nums.length;
for (int i = 0; i < nums.length; i++)
nums[i] = (i & 1) == 0 ? temp[--s] : temp[--t];
}
private int findMedium(int[] nums, int L, int R, int k) {
if (L >= R) return nums[R];
int i = partition(nums, L, R);
int cnt = i - L + 1;
if (cnt == k) return nums[i];
return cnt > k ? findMedium(nums, L, i - 1, k) : findMedium(nums, i + 1, R, k - cnt);
}
private int partition(int[] nums, int L, int R) {
int val = nums[L];
int i = L, j = R + 1;
while (true) {
while (++i < R && nums[i] < val) ;
while (--j > L && nums[j] > val) ;
if (i >= j) break;
swap(nums, i, j);
}
swap(nums, L, j);
return j;
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}