[LeetCode] 1144. Decrease Elements To Make Array Zigzag 递减元素使数组呈锯齿状

Posted 2021-06-21 Grandyang

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

这道题说是每次可以给数组中的任意数字减小1，现在想将数组变为之字形，就是数字大和小交替出现，有两种，一种是偶数坐标的数字均大于其相邻两个位置的数字，一种是奇数坐标的数字均大于其相邻的两个位置的数字。对于第一种情况来说，其奇数坐标位置的数字就均小于其相邻两个位置的数字，同理，对于第二种情况，其偶数坐标位置的数字就均小于其相邻两个位置的数字。这里我们可以分两种情况来统计减少次数，一种是减小所有奇数坐标上的数字，另一种是减小所有偶数坐标上的数字。减小的方法是找到相邻的两个数字中的较小那个，然后比其小1即可，即可用 nums[i] - min(left, right) + 1 来得到，若得到了个负数，说明当前数字已经比左右的数字小了，不需要再减小了，所以需要跟0比较，取较大值。这里用了一个大小为2的 res 数组，这用直接根据当前坐标i，通过 i%2 就可以更新对应的次数了，最终取二者中的较小值返回即可，参见代码如下：

class Solution {
public:
    int movesToMakeZigzag(vector<int>& nums) {
        int n = nums.size(), res[2] = {0, 0};
        for (int i = 0; i < n; ++i) {
            int left = i > 0 ? nums[i - 1] : 1001;
            int right = i < n - 1 ? nums[i + 1] : 1001;
            res[i % 2] += max(0, nums[i] - min(left, right) + 1);
        }
        return min(res[0], res[1]);
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1144

类似题目：

ZigZag Conversion

Binary Tree Zigzag Level Order Traversal

参考资料：

https://leetcode.com/problems/decrease-elements-to-make-array-zigzag/

https://leetcode.com/problems/decrease-elements-to-make-array-zigzag/discuss/350576/JavaC%2B%2BPython-Easy-and-concise

https://leetcode.com/problems/decrease-elements-to-make-array-zigzag/discuss/351306/C%2B%2B-O(n)-Brute-force-Easy-to-understand-and-detailed-explanation

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