Leetcode27. Remove Element

ddddLeetcode刷题

• Author：李英民 | Henry
• E-mail: li_yingmin@outlookdotcom
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• 转载请保留上述引用内容，谢谢配合！

Given an array and a value, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.

说明：

In computer science, an in-place algorithm is an algorithm which transforms input using no auxiliary data structure.

我的代码1：

//Author: Li_Yingmin@https://liyingmin.wixsite.com/henry
//Email: li_yingmin@outlook DOT com
//Leetcode-27
//T(n)=O(n), S(n)=O(1)
class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        if(nums.size()<1) return 0;
        for(auto it = nums.begin(); it != nums.end(); ++it)
        {
            if(*it == val)
            {
                nums.erase(it);
                --it;
            }
        }
        return nums.size();
    }
};

优秀代码1：

// LeetCode, Remove Element
// T(n)=O(n),S(n)=O(1)
class Soluntion
{
public:
    int removeElement(vector<int>& nums, int val)
    {
        return distance(nums.begin(), remove(nums.begin(), nums.end(), val));
    }
};

• remove接口，跟优秀代码2的实现几乎一样。
• Remove C++11

优秀代码2：

class Solution
{
public:
    int removeElement(vector<int>& nums, int target)
    {
        int index = 0;
        for (int i = 0; i < nums.size(); ++i)
            {
                if (nums[i] != target)
                    {
                        nums[index++] = nums[i];
                    }
            }
        return index;
    }
};