leetcode26. Remove Duplicates from Sorted Array

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本文介绍了一种在原地去除已排序数组中重复元素的方法，并确保每个元素只出现一次。通过两个示例展示了如何使用这种方法更新数组长度并保留唯一元素。同时提供了一种简洁的实现方式。

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题目描述：

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]);
}

My solution:

思路即为用一个指针来指示存入不重复元素的位置，每次遇到一个不重复的元素，就将该元素放到该指针处。

思路局限：代码冗余，看看别人是怎么写的！！！

int removeDuplicates(vector<int>& nums) {
        if(nums.empty()) return 0;
        int cur = 0; 
        for(int i = 0; i < nums.size(); i++){
            nums[cur] = nums[i];
            while(i < nums.size() - 1 && nums[i+1] == nums[i]){
                i++;
            }
            cur++;
        }
        return cur;
    }

Elegant implementation:

Author: leetcode ID: StefanPochmann

int removeDuplicates(vector<int>& nums) {
    int i = 0;
    for (int n : nums)
        if (!i || n > nums[i-1])
            nums[i++] = n;
    return i;
}

or:
int removeDuplicates(vector<int>& nums) {
    int i = !nums.empty();
    for (int n : nums)
        if (n > nums[i-1])
            nums[i++] = n;
    return i;
}