QUESTION:
You are given a 0-indexed array nums of size n consisting of non-negative integers.
You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:
If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation. After performing all the operations, shift all the 0’s to the end of the array.
For example, the array [1,0,2,0,0,1] after shifting all its 0’s to the end, is [1,2,1,0,0,0]. Return the resulting array.
Note that the operations are applied sequentially, not all at once.
Example 1:
Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].
Example 2:
Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.
Constraints:
2 <= nums.length <= 2000
0 <= nums[i] <= 1000
EXPLANATION:
for循环进行第一步判断,如果当前位置等于下一个位置, 那么就将当前位置翻倍, 后面位置置为0. 紧接着进行第二步, 如果当前数字不为0, 那么就加入到result中. 这样循环到结尾. 最后, 在末尾补上剩余的0即可.
SOLUTION:
class Solution {
func applyOperations(_ nums: [Int]) -> [Int] {
var nums = nums
var result: [Int] = []
for index in 0..<nums.count {
if (index + 1 != nums.count) {
if (nums[index] == nums[index + 1]) {
nums[index] += nums[index]
nums[index+1] = 0
}
}
if (nums[index] != 0) {
result.append(nums[index])
}
}
result.append(contentsOf: Array(repeating: 0, count: nums.count - result.count))
return result
}
}