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Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....
nums
nums[0] < nums[1] > nums[2] < nums[3]...
Example 1:
Input: nums = [1, 5, 1, 1, 6, 4] Output: One possible answer is [1, 4, 1, 5, 1, 6].
Example 2:
Input: nums = [1, 3, 2, 2, 3, 1] Output: 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?
这道题给了我们一个无序数组,让我们排序成摆动数组,满足nums[0] < nums[1] > nums[2] < nums[3]...,并给了我们例子。我们可以先给数组排序,然后在做调整。调整的方法是找到数组的中间的数,相当于把有序数组从中间分成两部分,然后从前半段的末尾取一个,在从后半的末尾去一个,这样保证了第一个数小于第二个数,然后从前半段取倒数第二个,从后半段取倒数第二个,这保证了第二个数大于第三个数,且第三个数小于第四个数,以此类推直至都取完,参见代码如下:
解法一:
// O(n) space class Solution { public: void wiggleSort(vector<int>& nums) { vector<int> tmp = nums; int n = nums.size(), k = (n + 1) / 2, j = n; sort(tmp.begin(), tmp.end()); for (int i = 0; i < n; ++i) { nums[i] = i & 1 ? tmp[--j] : tmp[--k]; } } };
这道题的Follow up让我们用O(n)的时间复杂度和O(1)的空间复杂度,这个真的比较难,参见网友的解答,(未完待续。。)
解法二:
// O(1) space class Solution { public: void wiggleSort(vector<int>& nums) { #define A(i) nums[(1 + 2 * i) % (n | 1)] int n = nums.size(), i = 0, j = 0, k = n - 1; auto midptr = nums.begin() + n / 2; nth_element(nums.begin(), midptr, nums.end()); int mid = *midptr; while (j <= k) { if (A(j) > mid) swap(A(i++), A(j++)); else if (A(j) < mid) swap(A(j), A(k--)); else ++j; } } };
类似题目:
Sort Colors
Kth Largest Element in an Array
Wiggle Sort
参考资料:
https://leetcode.com/problemset/algorithms/
https://leetcode.com/problems/wiggle-sort-ii/discuss/77706/Short-simple-C%2B%2B
https://leetcode.com/problems/wiggle-sort-ii/discuss/77677/O(n)%2BO(1)-after-median-Virtual-Indexing
https://leetcode.com/problems/wiggle-sort-ii/discuss/77682/Step-by-step-explanation-of-index-mapping-in-Java
https://leetcode.com/problems/wiggle-sort-ii/discuss/77681/O(n)-time-O(1)-space-solution-with-detail-explanations
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered:
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Given an unsorted array
nums
, reorder it such thatnums[0] < nums[1] > nums[2] < nums[3]...
.Example 1:
Example 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?
这道题给了我们一个无序数组,让我们排序成摆动数组,满足nums[0] < nums[1] > nums[2] < nums[3]...,并给了我们例子。我们可以先给数组排序,然后在做调整。调整的方法是找到数组的中间的数,相当于把有序数组从中间分成两部分,然后从前半段的末尾取一个,在从后半的末尾去一个,这样保证了第一个数小于第二个数,然后从前半段取倒数第二个,从后半段取倒数第二个,这保证了第二个数大于第三个数,且第三个数小于第四个数,以此类推直至都取完,参见代码如下:
解法一:
这道题的Follow up让我们用O(n)的时间复杂度和O(1)的空间复杂度,这个真的比较难,参见网友的解答,(未完待续。。)
解法二:
类似题目:
Sort Colors
Kth Largest Element in an Array
Wiggle Sort
参考资料:
https://leetcode.com/problemset/algorithms/
https://leetcode.com/problems/wiggle-sort-ii/discuss/77706/Short-simple-C%2B%2B
https://leetcode.com/problems/wiggle-sort-ii/discuss/77677/O(n)%2BO(1)-after-median-Virtual-Indexing
https://leetcode.com/problems/wiggle-sort-ii/discuss/77682/Step-by-step-explanation-of-index-mapping-in-Java
https://leetcode.com/problems/wiggle-sort-ii/discuss/77681/O(n)-time-O(1)-space-solution-with-detail-explanations
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered: