Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.

Example 1:

Input: S = "loveleetcode", C = 'e'
 Output:

Note:

S string length is in [1, 10000].
 C is a single character, and guaranteed to be in string S.
 All letters in S and C are lowercase. 

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int calmin(int a,int b)
{
    if(a < b)
        return a;
    else
        return b;
}

int calabs(int a,int b)
{
    if(a > b)
        return a-b;
    else
        return b-a;
}

int* shortestToChar(char* S, char C, int* returnSize) 
{
    int len = strlen(S);
    *returnSize = len;
    int *res = (int *)malloc(sizeof(int) * len);
    
    /*初始化*/
    for(int i = 0; i < len;i++)
    {
        res[i] = INT_MAX;
    }
    
    int pos = -len;
    
    for (int i = 0; i < len; ++i) 
    {
        if (S[i] == C) 
            pos = i;
        
        res[i] = calmin(res[i], calabs(i,pos));
    }
        
    
    for (int i = len - 1; i >= 0; --i) 
    {
        if (S[i] == C)  pos = i;
        res[i] = calmin(res[i], calabs(i,pos));
    }
    
    return res;
}

在编写这道题目的代码时，对memset函数差点使用错，在C++等实现方式中只需要两个循环就可以解决这个问题，但是C中，如果我对res全1初始化，那么数组中每个元素的值将是-1。这点与C++中的一种用法不同，比如：
 vector res(n, n);
 那么res中的每个值都将赋值为n.