LeetCode：26. Remove duplicates from an ordered array

To give you one Ascending order Array of nums , Would you please In situ Delete duplicate elements , Make each element Only once , Returns the new length of the deleted array . Elemental Relative order It should be maintained Agreement .

Because the length of an array cannot be changed in some languages , So you have to put the result in an array nums The first part of . More formally , If there is... After deleting duplicates k Elements , that nums Before k An element should hold the final result .

Insert the final result into nums Before k Return to... After a position k .

Don't use extra space , You must be there. In situ Modify input array And using O(1) Complete with extra space .

Criteria for judging questions :

The system will use the following code to test your solution :

int[] nums = [...]; //  Input array 
int[] expectedNums = [...]; //  The expected answer with the correct length 

int k = removeDuplicates(nums); //  call 

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    
    assert nums[i] == expectedNums[i];
}

If all assertions pass , Then your solution will be adopt .

Example 1：

 Input ：nums = [1,1,2]
 Output ：2, nums = [1,2,_]
 explain ： Function should return the new length  2 , And the original array  nums  The first two elements of are modified to  1, 2 . You don't need to think about the elements in the array that follow the new length .

Example 2：

 Input ：nums = [0,0,1,1,1,2,2,3,3,4]
 Output ：5, nums = [0,1,2,3,4]
 explain ： Function should return the new length  5 ,  And the original array  nums  The first five elements of are modified to  0, 1, 2, 3, 4 . You don't need to think about the elements in the array that follow the new length .

Tips ：

• 0 <= nums.length <= 3 * 10^4
• -10^4 <= nums[i] <= 10^4
• nums Pressed Ascending array

---

Their thinking

1. Because the title requires modification in place , So use double pointers . First create a slow pointer i, Point to the first element of the array , Create another fast pointer j, Point to the second element of the array
2. Then traverse the fast pointer j： if nums[j] and nums[i] Unequal , The first will i Increasing 1, And then put nums[i] To change the value of nums[j]
3. Finally, return the new length of the deleted array i+1 that will do

Code

/** * @param {number[]} nums * @return {number} */
var removeDuplicates = function(nums) {
    
    if(!nums.length) return 0;
    let i = 0;
    for(let j = 1; j < nums.length; j++){
    
        if(nums[j] !== nums[i]){
    
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
};