LeetCode 902. Numbers At Most N Given Digit Set

应该是常数 N的位数时间级别

个人这个方法超时很严重。。。可是特此记录 费劲巴拉写的。。。

超时：

int atMostNGivenDigitSet(char** D, int DSize, int N) {
    int *array = (int *)malloc(sizeof(int) * 10);
    for (int i = 0; i < 10; i ++)
        array[i] = 0;
    
    for (int j = 1; j < 10; j ++) {
        
        
        if (j - 1 < DSize) {
            char cc = D[j - 1][0];
            int c = cc - 48;
            array[c] = c;
        }
        //printf("\n%d",array[j]);
    }
    int totalSum = 0;
    for (int i = 1; i <= N; i ++) {
        
        long sum = 0;
        int num =  0;
        int x = i;
        int isFlag = 1;
        while ( x!= 0) {    
            isFlag = 0;
            num = x % 10;  //末尾数字
            sum = sum * 10;//进位
            sum += num;
            x = x / 10;
            if (array[num] == 0) {
                isFlag = 1;
                break;
            }
        }
        if (isFlag == 0) {
            totalSum ++;
        }
    }
    free(array);
    return totalSum;
}

int main(int argc, const char * argv[]) {
    @autoreleasepool {
        // insert code here...
        NSLog(@"Hello, World!\n");
        //[-2,-1] --->  -1
     
        char *s[] = {"1","3","5","7"};
        int maxNum = atMostNGivenDigitSet(s, 4, 100);
        printf("\n%d\n",maxNum);
    }
    return 0;
}