Leetcode 283) Move Zeroes

17 May 2021 in Algorithms

• The new position for zero was a bit unclear. but now it is clear! BUT, next zero position will be curzero+1 ! since we are going from left to right, there would be two cases 1. 000231 (0 is next to 0) 2. 0223400 (0 is not next to 0) => in this case, we swap with non zero value. and that is right next to !!!!

class Solution {
    public void moveZeroes(int[] nums) {
        int posZero=0;
        for(int i =0;i<nums.length;i++){
            /*
            if nums[i]==0 =>posZero won't be updated. because, it should hold the first 0 value of the array.
            we process from the beginning 
            */ 
            if(nums[i]!=0){
                //since we processed all 0 in front of 
                swap(posZero,i,nums);
                posZero++;
            }
        }

        /*
        Since we can simply override, the elemnts!!! I don't have to use while loop to implement the algorithm
        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                int ptr = i;
                while(ptr<nums.length && nums[ptr]==0){
                    ptr++;
                }
                if(ptr<nums.length)swap(i,ptr,nums);

            }
        }*/

    }

    public void swap(int i ,int j , int[] nums){
        int tmp = nums[i];
        nums[i]=nums[j];
        nums[j]=tmp;
    }
}

다른 풀이

• 등장한 0의 개수를 센 후에, index-등장한 0의 개수 를 해서 값을 저장해준다!

class Solution {
    public void moveZeroes(int[] nums) {
        int numZeros = 0;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == 0) {
                numZeros++;
            } else {
                int temp = nums[i];
                nums[i] = 0;
                //zero나온 만큼 인덱스가 앞으로 가니까..
                nums[i - numZeros] = temp;
            }
        }
    }
}