leetcode 27 Remove Element

最新推荐文章于 2021-04-19 23:04:21 发布

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本文介绍了一种在数组中移除指定元素并返回新长度的算法。提供了多种编程语言实现方式，包括C++和Python，并给出了简洁高效的代码示例。

Remove Element Total Accepted: 60351 Total Submissions: 187833 My Submissions

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

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

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c++ 解决方案：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int n = nums.size();
        int i = 0;
        while( i < n ) { 
            if( nums[i] == val ) {
                swap(nums[i], nums[n-1]);
                n--;
            } else {
                i++;
            }
        }
        return n;
    }
};

int removeElement(vector<int>& nums, int val)
{
    vector<int>::iterator  itr = nums.begin();
    while (itr != nums.end())
    {
        if (*itr == val)
            itr = nums.erase(itr);
        else
            ++itr;
    }
    return nums.size();
}

int removeElement(int A[], int n, int elem) {
    int begin=0;
    for(int i=0;i<n;i++) if(A[i]!=elem) A[begin++]=A[i];
    return begin;
}

python解决方案：

class Solution:
# @param    A       a list of integers
# @param    elem    an integer, value need to be removed
# @return an integer
def removeElement(self, A, elem):
    i = 0
    for j in range(len(A)):
        if A[j] != elem:
            A[i] = A[j]
            i += 1
    return i

史上最简洁的解决方案：

def removeElement(self, nums, val):
        nums[:] = [x for x in nums if x!=val]
        return len(nums)