Topic:
Given an array nums, write a function to move all zeros to the end of the array while preserving the relative order of the non-zero elements.
Example:
Input:,1,0,3,12 [0]
Output:,3,12,0,0 [1]
Copy the codeDescription:
- Must operate on the original array, cannot copy additional arrays.
- Minimize the number of operations.
The topic
They only limit operations on the original array, but they don’t limit changing the array length.
So the easiest way to do that is to find a 0 and delete it from the array and then append a 0 to the end of the array, and that’s what they’re asking for
/ * *
* @param {number[]} nums
* @return {void} Do not return anything, modify nums in-place instead.
* /
var moveZeroes = function(nums) {
let index = 0.
zoreNum = 0
while (index < nums.length - zoreNum) {
if (nums[index] === 0) {
nums.splice(index, 1)
zoreNum++
} else {
index++
}
}
}
Copy the codeDouble pointer
Declare two Pointers:
- A pointer left remains in the so-called position of the first 0 from front to back
- In addition, the pointer right is kept in an array position from front to back 0
- Substitution process: Replace the first encountered 0 with subsequent arrays one by one (update the index of the first 0)
var moveZeroes = function(nums) {
let len = nums.length,
left = 0.
right = 0;
while (right < len) {
if (nums[right]) {
[nums[left], nums[right]] = [nums[right], nums[left]]
left++
}
right++
}
}
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