LeetCode每日一题(2216. Minimum Deletions to Make Array Beautiful)

You are given a 0-indexed integer array nums. The array nums is beautiful if:

nums.length is even.
nums[i] != nums[i + 1] for all i % 2 == 0.
Note that an empty array is considered beautiful.

You can delete any number of elements from nums. When you delete an element, all the elements to the right of the deleted element will be shifted one unit to the left to fill the gap created and all the elements to the left of the deleted element will remain unchanged.

Return the minimum number of elements to delete from nums to make it beautiful.

Example 1:

Input: nums = [1,1,2,3,5]
Output: 1

Explanation: You can delete either nums[0] or nums[1] to make nums = [1,2,3,5] which is beautiful. It can be proven you need at least 1 deletion to make nums beautiful.

Example 2:

Input: nums = [1,1,2,2,3,3]
Output: 2

Explanation: You can delete nums[0] and nums[5] to make nums = [1,2,2,3] which is beautiful. It can be proven you need at least 2 deletions to make nums beautiful.

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105

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假设 i % 2 == 0, 如果 nums[i] != nums[i+1]，我们就可以检查下一个 pair， i += 2, 如果 nums[i] == nums[i+1], 此时我们可以删掉 nums[i+1]之前的任意一个数字(包括 nums[n+1])来保持 beautiful, 同时 nums[n+1]之后的数字奇偶位置都发生变化， 所以我们如果再 i += 2 的话就跳到了奇数位上， 所以此时我们需要 i += 1。 遍历完成后我们还需要看一下最终剩下的元素数量， 如果是奇数的话，为了满足题目中的第一个条件， 我们需要再删掉一个数

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impl Solution {
    pub fn min_deletion(nums: Vec<i32>) -> i32 {
        let mut i = 0;
        let mut deleted = 0;
        while i + 1 < nums.len() {
            if nums[i] == nums[i + 1] {
                deleted += 1;
                i += 1;
                continue;
            }
            i += 2
        }
        if (nums.len() - deleted as usize) % 2 == 0 {
            deleted
        } else {
            deleted + 1
        }
    }
}