【leetcode】1144. Decrease Elements To Make Array Zigzag

题目如下：

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

解题思路：本题无外乎两种情况，一种是nums[0] > nums[1]，另一种是nums[0] < nums[1]，把这两种情况计算一遍求较小值即可。在计算过程中，如果要求nums[i] > nums[i-1]，那么把nums[i-1] 减到nums[i] - 1；如果要求nums[i] < nums[i-1]，则把nums[i]减少到nums[i-1] - 1。

代码如下：

class Solution(object):
    def movesToMakeZigzag(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # flag -  0: decrease; 1:increase
        def process(flag,nums):
            count = 0
            for i in range(1, len(nums)):
                if flag == 0 and nums[i] >= nums[i - 1]:
                    count += (nums[i] - nums[i - 1] + 1)
                    nums[i] = nums[i - 1] - 1
                elif flag == 1 and nums[i] <= nums[i - 1]:
                    count += (nums[i - 1] - nums[i] + 1)
                    nums[i - 1] = nums[i] - 1
                flag = not flag
            return count


        #nums[0] > nums[1]
        res = process(0,nums[::])
        # nums[1] > nums[0]
        res = min(res,process(1,nums))
        return res

【leetcode】1144. Decrease Elements To Make Array Zigzag

原文：https://www.cnblogs.com/seyjs/p/11383437.html