1144-递减元素使数组呈锯齿状(Decrease Elements To Make Array Zigzag)

原文链接: https://leetcode-cn.com/problems/decrease-elements-to-make-array-zigzag

英文原文

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

中文题目

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

通过代码

高赞题解

[]
class Solution {
public:
    int movesToMakeZigzag(vector<int>& nums) {
        if (nums.size() < 2) return 0;
        int s1 = 0;
        int s2 = 0;
        int N = nums.size();
        for (int i = 0; i < N; ++i) {
            int l = (i > 0) ? nums[i - 1] : INT_MAX;
            int r = (i < N - 1) ? nums[i + 1] : INT_MAX;
            if (i & 1) {
                s1 += max(0, nums[i] - min(l, r) + 1);
            } else {
                s2 += max(0, nums[i] - min(l, r) + 1);
            }
        }
        return min(s1, s2);
    }
};