leetcode week17
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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.
问题描述:给定一个从小到大顺序排列的数组,要求返回非重复数字的数量n,并且将数组前n个数字替换位那些不重复的数字。
问题求解:设置两个指针sp,cp。初始条件为两者均指向index 0的位置。
cp指向的是不重复新元素替换的位置,而cp-1指向的就是前一个不重复元素。
sp我们要让其指向下一个新元素。
具体算法如下,结合算法理解。
class Solution {public: int removeDuplicates(vector<int>& nums) { int cp=0; for(int sp=0; sp<nums.size(); sp++){ if(sp==0||nums[cp-1]!=nums[sp]){ nums[cp++] = nums[sp]; } } return cp; }};
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