Writing notes day2

  Leetcode    1089 Copy Zero

Idea: double pointer

Step 1: use the pointer cur to record the last digit to be copied, and use the pointer dest to point to the position where the last digit is to be copied;

Implementation process: at the beginning, cur points to 0, and dest points to -1. When arr[cur] != 0, dest takes two steps, otherwise it takes one step; determine whether dest is already at the end of the array or has crossed the boundary;

Step 2: If the array is out of bounds, let arr[arr.size()-1] = 0; dest -= 2; cur --;

Step 3: Complete the copying from the back to the front according to the requirements.

class Solution {
public:
    void duplicateZeros(vector<int>& arr) {
        int cur = 0 , dest = -1;
        //确定要复习的最后一位数，使用cur指针指向
        while(cur < arr.size())
        {
            if(arr[cur] !=0)
            {
                dest++;
            }else{
                dest +=2;
            }
            //当dest指针指向数组最后一个位置或者越界一位，跳出循环，不再让cur 往后走
            if(dest >= arr.size()-1)
            {
                break;
            }
            ++cur;
        }
        // 如果dest处于arr.size()的位置
        if(dest == arr.size())
        {
            arr[arr.size()-1] = 0;
            dest -=2;
            --cur;
        }
        // 从右往左依次完成复
        while(cur >=0)
        {
            if(arr[cur])
            {
                arr[dest--] = arr[cur--];
            }else{
                arr[dest--] = 0;
                arr[dest--] = 0;
                cur--;
            }
        }
    }
};