Leetcode 26 Remove Duplicates from Sorted Array
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题目要求:
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.
要求去除数组中的重复元素,不能额外使用存储空间。刚开始使用3次遍历算法来做,而且用了iterator迭代器,但是超时了,而且算法效率不高,后来想了一下先将重复元素设置为0,然后在遍历数组,将值为0 的元素去除,但是也出现了一些问题,想来这种思路本身不好。参考了一篇博客的方法,使用i和index一起遍历,遇见相同的元素,就将i继续加,遇见不同的元素,就将index++,然后将nums[index]的位置值改为nums[i]的,结果很快就ac了。
代码如下:
class Solution {public: int removeDuplicates(vector<int>& nums) { if(nums.size() == 0) return 0; int index = 0; for(int i = 1; i < nums.size(); i++) { if(nums[index] != nums[i]) nums[++index] = nums[i]; } return index + 1; }};
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