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This article mainly introduces LeetCode how to delete the duplicates in the sorted array, which has a certain reference value, interested friends can refer to, I hope you can learn a lot after reading this article, the following let the editor take you to understand.

Meaning of the title

Given a sorted array, you need to delete the repeating elements in place so that each element appears only once, returning the new length of the removed array. Do not use extra array space, you must modify the input array in place and do it using O (1) extra space.

Sample

Example 1: the function should return the new length 2 given the array nums = [1Magne 1Mague 2], and the first two elements of the original array nums have been modified to 1Jing 2. You don't need to consider the elements in the array that exceed the new length.

Example 2: given nums = [0rect 0rect 1jiggle 1jiggle 2je 2rec 3je 3je 4], the function should return the new length 5, and the first five elements of the original array nums are modified to 0meme 1,2jol 3,4. You don't need to consider the elements in the array that exceed the new length.

After the problem-solving array is sorted, we can place two pointers I and j, where I is the slow pointer and j is the fast pointer. As long as nums [I] = nums [j], we increase j to skip duplicates. When we encounter nums [j] ≠ nums [I], the operation of skipping duplicates is over, so we must copy the value of its (nums [j]) to nums [I + 1]. Then increment I, and then we will repeat the same process again until j reaches the end of the array. Time complexity: O (n), assuming that the length of the array is n, then I and j traverse n steps at most. Space complexity: O (1).

Public int removeDuplicates (int [] nums) {if (nums.length = = 0) return 0; int I = 0; for (int j = 1; j < nums.length; jacks +) {if (nums [j]! = nums [I]) {icalendar; nums [I] = nums [j];}} return I + 1;}

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