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  • leetcode 27 Remove Element

    
    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)



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  • 原文地址:https://www.cnblogs.com/wuyida/p/6301327.html
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