LeetCode 26. 删除有序数组中的重复项

26. Remove Duplicates from Sorted Array

Given an integer array?nums?sorted in?non-decreasing order, remove the duplicates?in-place?such that each unique element appears only?once. The?relative order?of the elements should be kept the?same. Then return?the number of unique elements in?nums.

Consider the number of unique elements of?nums?to be?k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
• Return k.

Custom Judge:

The judge will test your solution with the following code:

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

int k = removeDuplicates(nums); // Calls your implementation

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 accepted.

Example 1:

Input: nums = [1,1,2]

Output: 2, nums = [1,2,_]

Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]

Output: 5, nums = [0,1,2,3,4,,,,,_]

Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

1 -100

nums is sorted in non-decreasing order.

?

解题思路:

1、双指针

2、索引 index

法一：?

class Solution {
    public int removeDuplicates(int[] nums) {
        // O(n)
        int n = nums.length;
        int slow = 0;
        int fast = 1;
        if (nums == null || n == 0) return 0;
        while (fast < n) {
            if (nums[slow] != nums[fast]) {
                nums[slow + 1] = nums[fast];
                ++slow;
            }
            ++fast;
        }
        return slow + 1;
     }
 }

?法二：

class Solution {
    public int removeDuplicates(int[] nums) {
        // O(n)
        int n = nums.length;
        if (n < 2) return n;
        int j = 1;
        for (int i = 1; i < n; i++) {
            if (nums[i] != nums[i-1]) {
                nums[j++] = nums[i];
            }
        }
        return j;
    }
}