leetcode 26 Remove Duplicates from Sorted Array

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 发布

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文章标签：数据结构与算法 python

原文链接：http://www.cnblogs.com/blfbuaa/p/6719492.html

本文介绍了如何在不使用额外空间的情况下从已排序数组中去除重复元素，并保持原有数组顺序不变的方法。通过多种编程语言实现，包括C++及Python，展示了不同的算法思路，如迭代、STL函数应用等。

Remove Duplicates from Sorted ArrayTotal Accepted: 66627 Total Submissions: 212739 My Submissions

Given a sorted array, remove the duplicates in place such that each element appear onlyonce 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 ofnums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length

64ms好像有点慢了。这种方法有点讨巧了，绕过了算法的部分。以后还是少写这种代码，多练习算法

我的解决方式：

class Solution {
public:
    int removeDuplicates(vector<int>& nums)
    {
        set<int> result;
        for(int i = 0;i< nums.size();i++)
        {
            result.insert(nums[i]);
        }
        nums.clear();
        set<int>::iterator iter = result.begin();
        for(;iter!=result.end();iter++)
        {
            nums.push_back(*iter);
        }
        return nums.size();
    }
};

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int start=1,N = nums.size();
        if(N<=1) return N;
        for(int i=1;i<nums.size();i++) {
            if(nums[i]!=nums[i-1]){
                nums[start]=nums[i];
                start++;
            }
        }
        return start;
    }
};

一行代码的STL:

class Solution { public: int removeDuplicates(int A[], int n) { return distance(A, unique(A, A+n)); } };

int removeDuplicates(vector<int>& nums) {
    if(nums.size() <= 1) return nums.size();

    vector<int>::iterator it1,it2;
    for(it1=nums.begin(),it2=nums.begin()+1; it2 != nums.end();) {
        if(*it2 == *it1) it2=nums.erase(it2);
        else {it1++;it2++;}
    }

    return nums.size();
}

python解决方式：

class Solution:
    # @param a list of integers
    # @return an integer
    def removeDuplicates(self, A):
        if not A:
            return 0

        newTail = 0

        for i in range(1, len(A)):
            if A[i] != A[newTail]:
                newTail += 1
                A[newTail] = A[i]

        return newTail + 1

转载于:https://www.cnblogs.com/blfbuaa/p/6719492.html