leetcode 26 Remove Duplicates from Sorted Array-优快云博客

本文提供了一种在原地且使用常数内存的方式去除已排序数组中重复元素的方法，并给出了C++及Python的实现代码。该方法通过遍历数组并仅保留不重复的元素来工作，最终返回不含重复项的新数组长度。

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

转载于:https://www.cnblogs.com/wangyaning/p/7853957.html