[LeetCode] 2422. Merge Operations to Turn Array Into a Palindrome

You are given an array nums consisting of positive integers.

You can perform the following operation on the array any number of times:

• Choose any two adjacent elements and replace them with their sum.
  • For example, if nums = [1,2,3,1], you can apply one operation to make it [1,5,1].

Return the minimum number of operations needed to turn the array into a palindrome.

Example 1:

Input: nums = [4,3,2,1,2,3,1]
Output: 2
Explanation: We can turn the array into a palindrome in 2 operations as follows:
- Apply the operation on the fourth and fifth element of the array, nums becomes equal to [4,3,2,3,3,1].
- Apply the operation on the fifth and sixth element of the array, nums becomes equal to [4,3,2,3,4].
The array [4,3,2,3,4] is a palindrome.
It can be shown that 2 is the minimum number of operations needed.

Example 2:

Input: nums = [1,2,3,4]
Output: 3
Explanation: We do the operation 3 times in any position, we obtain the array [10] at the end which is a palindrome.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 106

使用合并操作将数组转换为回文序列。

给定一个由 正整数 组成的数组 nums。

可以对阵列执行如下操作，次数不限:

选择任意两个 相邻 的元素并用它们的 和 替换 它们。
例如，如果 nums = [1,2,3,1]，则可以应用一个操作使其变为 [1,5,1]。
返回将数组转换为 回文序列 所需的 最小 操作数。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/merge-operations-to-turn-array-into-a-palindrome
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思路是贪心 + 双指针。从数组的左右两边开始看，

• 如果 nums[left] == nums[right]，两边数字一样，left++, right--，继续比较中间的元素
• 如果 nums[left] != nums[right]，两边数字不一样，此时需要合并，谁较小就让谁与其邻居元素合并，并移动指针
• 当 left 指针与 right 指针相遇的时候，返回一共合并了几次

时间O(n)

空间O(1)

Java实现

class Solution {
    public int minimumOperations(int[] nums) {
        int res = 0;
        int left = 0;
        int right = nums.length - 1;
        while (left < right) {
            if (nums[left] < nums[right]) {
                nums[left + 1] += nums[left];
                left++;
                res++;
            } else if (nums[left] > nums[right]) {
                nums[right - 1] += nums[right];
                right--;
                res++;
            } else {
                left++;
                right--;
            }
        }
        return res;
    }
}

 

LeetCode 题目总结