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 multiplynums[i]
by2
and setnums[i + 1]
to0
. 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 its0
'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.
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].
Input: nums = [0,1] Output: [1,0] Explanation: No operation can be applied, we just shift the 0 to the end.
2 <= nums.length <= 2000
0 <= nums[i] <= 1000
impl Solution {
pub fn apply_operations(nums: Vec<i32>) -> Vec<i32> {
let mut nums = nums;
let mut i = 0;
let mut ret = vec![0; nums.len()];
for j in 0..nums.len() {
if j + 1 < nums.len() && nums[j] == nums[j + 1] {
nums[j] += nums[j];
nums[j + 1] = 0;
}
if nums[j] != 0 {
ret[i] = nums[j];
i += 1;
}
}
ret
}
}