LeetCode solution summary 2460. Perform operations on arrays

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GitHub synchronous brushing project:

https://github.com/September26/java-algorithms

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describe:

Given an   array with subscripts starting  at 0nums  , the size of the array is  n , and it consists of  non-negative  integers.

You need to perform  n - 1 a step operation on the array, where the step  i operation (   counting  from 0 ) requires  the following instructions to be executed nums on the first  element:i

• If  nums[i] == nums[i + 1] , then  nums[i] the value of , becomes the original  2 multiple, nums[i + 1] and the value of  0 . Otherwise, skip this step.

After all operations are performed  , 0 move  all   to  the end of the array  .

• For example, the array  [1,0,2,0,0,1] will become after all  0 shifted to the end  [1,2,1,0,0,0] .

Returns an array of results.

Note that  operations should  be executed sequentially  , not all at once.

Example 1:

Input: nums = [1,2,2,1,1,0]
 Output: [1,4,2,0,0,0]
 Explanation: Do the following:
- i = 0: nums[0] and nums[1] are not equal, skip this step.
- i = 1: nums[1] and nums[2] are equal, the value of nums[1] becomes twice the original value, and the value of nums[2] becomes 0. The array becomes [1, 4 , 0,1,1,0 ].
- i = 2: nums[2] and nums[3] are not equal, so skip this step.
- i = 3: nums[3] and nums[4] are equal, the value of nums[3] becomes twice the original value, and the value of nums[4] becomes 0. The array becomes [1,4,0, 2 , 0 ,0] .
- i = 4: nums[4] and nums[5] are equal, the value of nums[4] becomes twice the original value, and the value of nums[5] becomes 0. The array becomes [1,4,0,2, 0 , 0 ].
After performing all operations, move all 0's to the end of the array, resulting in an array [1,4,2,0,0,0].

Example 2:

Input: nums = [0,1]
 Output: [1,0]
 Explanation: Can't do anything except move 0 to the end.

hint:

• 2 <= nums.length <= 2000
• 0 <= nums[i] <= 1000

Problem-solving ideas:

/**

* First, perform corresponding processing for the value of nums[i] == nums[i + 1].

* Then move all 0s to the end of the array, we set an index, record the current number that is not 0, and exchange positions with num[index] every time a number that is not 0 is traversed.

*/

code:

vector<int> Solution2460::applyOperations(vector<int> &nums)
{
    for (int i = 0; i < nums.size() - 1; i++)
    {
        if (nums[i] != nums[i + 1])
        {
            continue;
        }
        nums[i] = nums[i] * 2;
        nums[i + 1] = 0;
    }
    int index = 0;
    for (int i = 0; i < nums.size(); i++)
    {
        if (nums[i] == 0)
        {
            continue;
        }
        int local = nums[index];
        nums[index] = nums[i];
        nums[i] = local;
        index++;
    }

    return nums;
};