26. Remove Duplicates from Sorted Array

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本文介绍了一个高效的算法，用于在不使用额外空间的情况下，从已排序的数组中去除重复元素，只保留唯一值，并返回这些唯一元素的数量。算法通过双指针技术实现，其中一个指针跟踪当前唯一元素的位置，另一个遍历整个数组，当遇到新的唯一元素时进行更新。这种方法确保了O(1)的额外空间复杂度。

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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]);

要求0(1)的空间复杂度，并且修改原数组，使其不同的元素都在数组前面。返回不同元素个数

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(!nums.size()){
            return 0;
        }
        int i=0,j=1;
        for(;j<nums.size();j++){
            if(nums[i]!=nums[j]){
                nums[++i]=nums[j];
            }
        }
        return i+1;
    }
};