Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return 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.

Example 1:

Given nums = [1,1,2],

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

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Analysis

Two Pointers 快慢指针:

两个指针，i，j：

i - 顺序存储下一个unique元素的index

j - 寻找下一个unique元素的index

主要逻辑是：

内层循环：只要nums[j] == nums[j + 1] 则 j++。若循环结束，说明j+1位置是下一个unique元素，而j位置是本次外层循环需要存的unique元素的index。j的额外说明：内层循环结束时，j是当前unique元素组的最后一个index。

外层循环：每次nums[i] = nums[j]，将nums[j]赋值给nums[i]，（0 ~ i代表排序好的unique元素）

两层循环结束时，i指向下一个待填入元素的index，因此实际0 ~ i - 1位置是我们所需元素，但是新数组总长度恰好也是 i - 1 + 1 = i

Just illustrates index changes, keeping original nums[] array as reference

-------------

1st inner loop ends

[1, 1, 2, 2, 3]
 ^  ^
 i  j

i = 0, j = 1


1st outer loop ends:

[1, 1, 2, 2, 3]
    ^  ^
    i  j

i = 1, j = 2

-------------

2nd inner loop ends:

[1, 1, 2, 2, 3]
    ^     ^
    i     j

i = 1, j = 3

2nd outer loop ends:

[1, 1, 2, 2, 3]
       ^     ^
       i     j

i = 2, j = 4

另一种Two Pointers实现，同样是slow, fast runner:

https://leetcode.com/articles/remove-duplicates-from-sorted-array/

Since the array is already sorted, we can keep two pointers i and j, where i is the slow-runner while j is the fast-runner. As long as nums[i] = nums[j], we increment j to skip the duplicate.

When we encounter nums[j] !​=nums[i], the duplicate run has ended so we must copy its value to nums[i + 1]. ii is then incremented and we repeat the same process again until j reaches the end of array.

外层循环 j (1 ~ nums.length - 1)，只要nums[j] != nums[i]就更新i为i++ ，并存入nums[i] = nums[j]

这种思路其实更简洁， input: [1, 1, 2, 2, 3]：

[1, 1, 2, 2, 3]
 ^  ^
 i  j


[1, 1, 2, 2, 3]
 ^     ^
 i     j

nums[j] != nums[i] 
nums[++i] = nums[j]


[1, 2, 2, 2, 3]
    ^  ^
    i  j


[1, 2, 2, 2, 3]
    ^     ^
    i     j


[1, 2, 2, 2, 3]
    ^        ^
    i        j

nums[j] != nums[i] 
nums[++i] = nums[j]


[1, 2, 3, 2, 3]
       ^     ^
       i     j

Loop ends, return i + 1

Solution

Two Pointers - O(n) time, O(1) space - each i, j traverse at most n steps

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

        if (nums == null || nums.length == 0) {
            return 0;
        }
        int n = nums.length;
        // i - the index to store unique element
        // j - the index that search next unique element
        int i = 0, j = 0;
        while (i < n && j < n) {
            // searching for next unique element
            while(j < n - 1 && nums[j] == nums[j + 1]) {
                j++;
            }
            // the next unique element found is (starting) with index j+1
            // current index for unique to be stored is j
            // the index to store unique element is i 
            nums[i] = nums[j];
            i++;
            j++;
        }
        return i;
    }
}

Another Two Pointers

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;
}