Daily question 1: remove elements

link : Remove elements

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The idea of this problem is relatively clear , We can use the double pointer method , But for the integer array, we can use the subscript to represent the pointer .

cur It is used to judge whether it is val, and tmp The function of is to count and compare with cur Data exchange at ：

	int cur = 0;
    int tmp = 0;

Ideas ：
The first thing you must do is loop through these two pointers , The cyclic condition is cur Less than numsSize（ Number of original arrays ）, So you can traverse the entire array .

And within the cycle , about cur, If the value at this subscript is val, Then let cur Go back to , Until you meet someone who is not val Of . And wait for cur It's not val It's about , let cur The data at is consistent with tmp Data exchange at , then tmp and cur All the way back , Continue to cycle , until cur After traversing the array .

Readers can construct according to this figure .
Code ：

int removeElement(int* nums, int numsSize, int val){
    
    int cur = 0;
    int tmp = 0;

    while(cur < numsSize)
    {
    
    	// If it is equal cur Gaga 
        if(nums[cur] == val)
        {
    
            cur++;
        }
        // In case of inequality, exchange the two , Then add them separately 
        else
        {
    
            nums[tmp] = nums[cur];
            tmp++;
            cur++;
        }
    }

    return tmp;

}