[LeetCode] 280. Wiggle Sort 摆动排序

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].

给一个没有排序的数组，将其重新排序成nums[0] <= nums[1] >= nums[2] <= nums[3]....的样子，要求in-place。

解法：遍历一遍数组, 如果是奇数位置并且其值比下一个大,则交换其值, 如果是偶数位置并且其值比下一个小, 则交换其值. 时间复杂度是O(N)。注意index和实际的位置差1，所以奇偶相反。

Java:

public class Solution {
    public void wiggleSort(int[] nums) {
        if (nums == null || nums.length < 2) return;
        for (int i = 1; i < nums.length; i++) {
            if ((i % 2 == 0 && nums[i] > nums[i - 1]) || (i % 2 == 1 && nums[i] < nums[i - 1])) {
                int tmp = nums[i];
                nums[i] = nums[i - 1];
                nums[i - 1] = tmp;
            } 
        }
    }
}　　

Java:

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] < nums[i - 1]) {
                    swap(nums, i);
                } 
            } else {
                if (nums[i] > nums[i - 1]) {
                    swap(nums, i);
                }
            }
        }
    }
    
    private void swap(int[] nums, int i) {
        int tmp = nums[i - 1];
        nums[i - 1] = nums[i];
        nums[i] = tmp;
    }
}　　

Python:

# Time:  O(n)
# Space: O(1)
class Solution(object):
    def wiggleSort(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        for i in xrange(1, len(nums)):
            if ((i % 2) and nums[i - 1] > nums[i]) or \
                (not (i % 2) and nums[i - 1] < nums[i]):
                # Swap unordered elements.
                nums[i - 1], nums[i] = nums[i], nums[i - 1]

C++:

// Time:  O(n)
// Space: O(1)
class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        for (int i = 1; i < nums.size(); ++i) {
            if (((i % 2) && nums[i] < nums[i - 1]) ||
                (!(i % 2) && nums[i] > nums[i - 1])) {
                // Swap unordered elements.
                swap(nums[i], nums[i - 1]);
            }
        }
    }
};　

C++:

// Time Complexity O(nlgn)
class Solution {
public:
    void wiggleSort(vector<int> &nums) {
        sort(nums.begin(), nums.end());
        if (nums.size() <= 2) return;
        for (int i = 2; i < nums.size(); i += 2) {
            swap(nums[i], nums[i - 1]);
        }
    }
};

C++:

// Time Complexity O(n)
class Solution {
public:
    void wiggleSort(vector<int> &nums) {
        if (nums.size() <= 1) return;
        for (int i = 1; i < nums.size(); ++i) {
            if ((i % 2 == 1 && nums[i] < nums[i - 1]) || (i % 2 == 0 && nums[i] > nums[i - 1])) {
                swap(nums[i], nums[i - 1]);
            }
        }
    }
};

　　

 

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[LeetCode] Wiggle Sort II 摆动排序 II

 

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