LeetCode

Remove Element

2013.12.7 04:33

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.

Solution1:

　　My first solution is scan the array from both ends, if a target value is found on the left side, then get a non-target value from the right side to fill this position. When left and right meets somewhere in the middle, the scan is done.

　　Time complexity is O(n),  space complexity is O(1).

Accepted code:

class Solution {
public:
    int removeElement(int A[], int n, int elem) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int i, j, k;
        int len;
        
        if(A == nullptr){
            return 0;
        }
        
        len = n;
        i = 0;
        j = n - 1;
        while(i <= j){ // BUG: j > i, fixed here
            if(A[i] == elem){
                while(j >= i){ // BUG: j > i, fixed here
                    if(A[j] == elem){
                        --j;
                        --len;
                    }else{
                        A[i] = A[j];
                        ++i;
                        --j;
                        --len;
                        break;
                    }
                }
            }else{
                ++i;
            }
        }
        
        return len;
    }
};

Solution2:

　　It was mentioned in the problem that "the order can be changed", but there seemed to be no extra benefit from changing the order. So let's do this without changing the order.

　　Scan the array from left to right, when ever a target value is found, record the accumulated number of appearance of the target, and use this as an $offset. Move other non-target value back by $offset positions to make a new array.

　　Time complexity is O(n), space complexity is O(1).

Accepted code:

class Solution {
public:
    int removeElement(int A[], int n, int elem) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int i, j;
        int len;
        
        if(A == nullptr){
            return 0;
        }
        
        len = n;
        i = 0;
        j = 0;
        while(i < n){
            if(A[i] == elem){
                --len;
                ++i;
            }else{
                A[j++] = A[i++];
            }
        }
        
        return len;
    }
};