LeetCode OJ 26 Remove Duplicates from Sorted Array
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毕设做累了,无聊刷个题压压惊。今天是第一次做leetcode的题目,做题顺序按照分类来做。
该题分类为线性表,题目难度为简单。
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.
分析:时间复杂度为O(n)时必然只需要扫描一遍数组中的数据。之后考虑空间复杂度,因为编程基础比较薄弱,最先想到的是再创建一个数组然后存放数据,现在这样空间复杂度会提高,由于题目中提到不在乎排序之后的数组里都有什么。现在可以在原数组上进行操作。解题思路如下:
设立两个tag,tag1从0开始,tag2从1开始,tag2为循环控制变量。当Array【tag1】与Array【tag2】相等时,tag2++,tag1不做任何操作;当不相等时,tag1++,之后Array【Tag1】=Array【Tag2】。
AC代码如下:
class Solution {public: int removeDuplicates(vector<int>& nums) { int length = nums.size(); if(length<=1) return length; int index=0; for(int i =1;i<length;i++){ if(nums[index]!=nums[i]) nums[++index]=nums[i]; } return index+1; }};算法性能分析如下:
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