leetcode 26 Remove Duplicates from Sorted Array

最新推荐文章于 2024-01-02 23:35:28 发布

原创最新推荐文章于 2024-01-02 23:35:28 发布 · 2.6k 阅读

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本文介绍了一种方法，在不使用额外空间的情况下，去除有序数组中的重复元素，并返回新的数组长度。提供了多种解决方案，包括使用STL、迭代器等技巧，以及一行代码的简洁实现。

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，这个方法有点讨巧了，绕过了算法的部分，以后还是少写这样的代码，多练习算法

我的解决方案：

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解决方案：

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