26. 从排序数组中删除重复项 - Java

Question

Question: Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

public int removeDuplicates(int[] nums) {
if (nums.length == 0) return 0;
int i = 0;
for (int j = 1; j < nums.length; j++) {
    if (nums[j] != nums[i]) {
        i++;
        nums[i] = nums[j];
    }
}
return i + 1;

}

What exactly does the return statement do here. What does return i + 1 mean here?

Answer 1

The return i + 1 is returning how many unique integers are there. I believe this is a Leetcode problem, and since its in place, the int[] is passed in by reference, Leetcode wants to know how many numbers to check (you're supposed to put the unique numbers in the first i + 1 spots).

If you look at the question, it says: Which means that you return the length of the array.

So, if you have the array [1,1,2,3,4,4] , you would turn that into [1,2,3,4,...] , where the ... is the rest of the array. However, you return 4 because the length of the new array should be 4 .

Hope this clears things up for you!

Answer 2

Your question has been already answered here ; in addition to that, we can also start from zero and remove the first if statement:

Test with a b.java file:

import java.util.*;

class Solution {
    public static final int removeDuplicates(
        final int[] nums
    ) {

        int i = 0;

        for (int num : nums)
            if (i == 0 || num > nums[i - 1]) {
                nums[i++] = num;
            }

        return i;
    }
}


class b {
    public static void main(String[] args) {
        System.out.println(new Solution().removeDuplicates(new int[] { 1, 1, 2}));
        System.out.println(new Solution().removeDuplicates(new int[] { 0, 0, 1, 1, 1, 2, 2, 3, 3, 4}));
    }
}

prints

2
5

Answer 3

• I tried in this easy way. Here Time complexity is O(n) and space complexity: O(1).

   static int removeDuplicates(int[] nums){ if(nums.length == 0) { return 0; } int value = nums[0]; int lastIndex = 0; int count = 1; for (int i = 1; i < nums.length; i++) { if(nums[i] > value) { value = nums[i]; lastIndex = lastIndex+1; nums[lastIndex] = value; count++; } } return count; }

Answer 4

class Solution {
public int removeDuplicates(int[] nums) {
    int n = nums.length;
    if (n == 0 || n == 1)
        return n;
    
    int j = 0;
    for (int i=0; i<n-1; i++)
            if (nums[i]!= nums[i+1])
                nums[j++] = nums[i];
            
        
    nums[j++]=nums[n-1];
    return j;
}

}

Answer 5

public class RemoveDuplicateSortedArray {
    //Remove Duplicates from Sorted Array
    public static void main(String[] args) {
        int[] intArray = new int[]{0, 0, 1, 1, 1, 2, 2, 3, 3, 4};
        int count = extracted(intArray);
        for (int i = 0; i < count; i++) {
            System.out.println(intArray[i]);
        }
    }

    private static int extracted(int[] intArray) {
        int size = intArray.length;
        int count = 1;
        if (size == 1) {
            return 1;
        } else if (size == 2) {
            if (intArray[0] == intArray[1]) {
                return 1;
            } else {
                return 2;
            }
        } else {
            for (int i = 0, j = i + 1; j < size; j++) {
                if (intArray[i] < intArray[j]) {
                    i++;
                    intArray[i] = intArray[j];
                    count++;
                }
            }
            return count;
        }
    }
}