Arrays | 27. Remove elements

zero, description

Topic link: 27. Remove elements

1. Idea:

First of all, I want to use two pointers, the left pointer goes from beginning to end, and the right goes from tail to head.
When the left pointer points to arr[ i ] != target, record the index L, and the
right pointer starts from the end to the beginning, when arr[ j ]==target, record the index R.

int removeElement(vector<int>& nums, int val) {
    
    
        int N=nums.size();
        int l=0,r=N-1;     //存储下一次循环的地点
        if(N==0) return {
    
    };
  
        for(int i=l;i<r;i++){
    
    
            if(nums[i]==val){
    
    
                l=i+1;       //留作下次开始
                for(int j=r;j>=i;j--){
    
    
                    if(nums[j]!=val){
    
    
                        r=j-1;
                        nums[i]=nums[j];
                        break;
                    }
                }
            }
        }
        return l+1;
    }

When inputting nums=[1] val=1, the correct result is [ ], but the result returned by the code is [1]

Two, reference

Code Caprice 27-Remove Elements

/**
* 相向双指针方法，基于元素顺序可以改变的题目描述改变了元素相对位置，确保了移动最少元素
* 时间复杂度：O(n)
* 空间复杂度：O(1)
*/
int removeElement(vector<int>& nums, int val) {
    
    
        int leftIndex = 0;
        int rightIndex = nums.size() - 1;
        
        while(leftIndex<=rightIndex){
    
    
            // 找左边等于val的元素
            while (leftIndex <= rightIndex && nums[leftIndex] != val){
    
    
                ++leftIndex;
            }
            //跳出while时候，leftIndex表示此时左边等于val的元素
            
            // 找右边不等于val的元素
            while (leftIndex <= rightIndex && nums[rightIndex] == val) {
    
    
                -- rightIndex;
            }
            //跳出while时候，leftIndex表示此时右边不等于val的元素
            
            // 将右边不等于val的元素覆盖左边等于val的元素
            if (leftIndex < rightIndex) {
    
    
                nums[leftIndex++] = nums[rightIndex--];
            }
        }
        return leftIndex;   // leftIndex一定指向了最终数组末尾的下一个元素
    }
};

example

nums = [3,2,2,3], val = 3
leftIndex=0, rightIndex=3
nums[ leftIndex=0 ] == val, leftIndex=0
nums[ rightIndex=3 ] == val, rightIndex=2
nums[ rightIndex=2]! = val Jump out of while
to the end if, at this time leftIndex=0, rightIndex=2 exchange to get nums = [2,2,3,3]

Summarize

• Use while instead of for
• The while implementation is also the function of stopping the left pointer in the if condition