★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ?微信公众号:山青咏芝(shanqingyongzhi) ?博客园地址:山青咏芝(https://www.cnblogs.com/strength
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?微信公众号:山青咏芝(shanqingyongzhi)
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Given an array nums
of integers, a move consists of choosing any element and decreasing it by 1.
An array A
is a zigzag array if either:
- Every even-indexed element is greater than adjacent elements, ie.
A[0] > A[1] < A[2] > A[3] < A[4] > ...
- OR, every odd-indexed element is greater than adjacent elements, ie.
A[0] < A[1] > A[2] < A[3] > A[4] < ...
Return the minimum number of moves to transform the given array nums
into a zigzag array.
Example 1:
Input: nums = [1,2,3] Output: 2 Explanation: We can decrease 2 to 0 or 3 to 1.
Example 2:
Input: nums = [9,6,1,6,2] Output: 4
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 1000
给你一个整数数组 nums
,每次 操作 会从中选择一个元素并 将该元素的值减少 1。
如果符合下列情况之一,则数组 A
就是 锯齿数组:
- 每个偶数索引对应的元素都大于相邻的元素,即
A[0] > A[1] < A[2] > A[3] < A[4] > ...
- 或者,每个奇数索引对应的元素都大于相邻的元素,即
A[0] < A[1] > A[2] < A[3] > A[4] < ...
返回将数组 nums
转换为锯齿数组所需的最小操作次数。
示例 1:
输入:nums = [1,2,3] 输出:2 解释:我们可以把 2 递减到 0,或把 3 递减到 1。
示例 2:
输入:nums = [9,6,1,6,2] 输出:4
提示:
1 <= nums.length <= 1000
1 <= nums[i] <= 1000
Runtime: 8 ms Memory Usage: 20.8 MB
1 class Solution { 2 func movesToMakeZigzag(_ nums: [Int]) -> Int { 3 var n:Int = nums.count 4 var j:Int = 0 5 var s:Int = 0 6 var t:Int = 0 7 for i in stride(from:0,to:n,by:2) 8 { 9 j = 0 10 if i != 0 11 { 12 j = max(j,nums[i]-nums[i-1]+1) 13 } 14 if i + 1 < n 15 { 16 j = max(j,nums[i]-nums[i+1]+1) 17 } 18 s += j 19 } 20 for i in stride(from:1,to:n,by:2) 21 { 22 j = 0 23 if i != 0 24 { 25 j = max(j,nums[i]-nums[i-1]+1) 26 } 27 if i + 1 < n 28 { 29 j = max(j,nums[i]-nums[i+1]+1) 30 } 31 t += j 32 } 33 return min(s,t) 34 } 35 }