Leetcode笔记整理—easy篇—Remove Duplicates from Sorted Array

本文探讨了一种在不使用额外空间的情况下从已排序数组中删除重复元素的方法，并提供了两种不同的实现方案。一种是通过元素覆盖的方式，另一种则是更高效的只使用一次遍历的解决方案。

Remove Duplicates from Sorted Array

HR，题目描述：
Given a sorted array, 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 in place with constant memory.

For example,
Given input array 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 new length.

这里写图片描述

还是不大行啊，不过这次是有预感的效率比较低，，，
实现：

int removeDuplicates(int* nums, int numsSize) {
    int length = 1;
    if(nums == NULL || (*nums == 0 && numsSize == 0))
    {
        length = 0;
        return length;
    }
    for(int i = 0 ; i < numsSize-1; i++)
    {
        if(nums[i] == nums[i+1])
        {
            for(int j = i; j < numsSize-1; j++)
            {
                nums[j] = nums[j + 1];
            }
            i--;
            numsSize--;
        }
        else
        length++;
    }
    return length;
}

貌似关键就在那个for里了，可以优化的一个方式就是去掉for，和之后的参考案例也有关，，

参考案例：

class Solution {
    public:
    int removeDuplicates(int A[], int n) {
        if(n < 2) return n;
        int id = 1;
        for(int i = 1; i < n; ++i) 
            if(A[i] != A[i-1]) A[id++] = A[i];
        return id;
    }
};

膜拜中。。。