Leetcode: Remove Element (4ms)

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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 Array Two Pointers

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updated:
move all val to right and all non-val to the left with two pointers:
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3ms
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class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int l=0, r=nums.size()-1;
        while(l<=r){
            while (nums[l]!=val && l<nums.size()){
                l++;
            }
            while (r>=l && r>=0 && nums[r]==val){
                r--;
            }
            if(r<l)break;
            nums[l]=nums[r];
            nums[r]=val;
            l++;
            r--;
        }
        return l;
       
       
    }
};

Analysis:

Sweep i from left to right, and delete the target in place directly. 

/////////////////////////////////////////

class Solution {

public:

    int removeElement(vector<int>& nums, int val) {

        int n=nums.size();

        vector<int>::iterator it=nums.begin();

        for (int i=0;i<n;){

            if(nums[i]==val){

                nums.erase(it+i);

                --i;

                n=nums.size();//should update the size after each removing 

            }

            ++i;

        }

        return nums.size();

    }

};