iOS 合并两个有序数组

- (void)merge {
    /*
     有序数组A：1、4、5、8、10...1000000，有序数组B：2、3、6、7、9...999998，A、B两个数组不相互重复，请合并成一个有序数组C，写出代码和时间复杂度。
     */
    //(1).
    NSMutableArray *A = [NSMutableArray arrayWithObjects:@4,@5,@8,@10,@15, nil];
//    NSMutableArray *B = [NSMutableArray arrayWithObjects:@2,@6,@7,@9,@11,@17,@18, nil];
    NSMutableArray *B = [NSMutableArray arrayWithObjects:@2,@6,@7,@9,@11,@12,@13, nil];
    NSMutableArray *C = [NSMutableArray array];
    int count = (int)A.count+(int)B.count;
    int index = 0;
    for (int i = 0; i < count; i++) {
        if (A[0]<B[0]) {
            [C addObject:A[0]];
            [A removeObject:A[0]];
        }
        else if (B[0]<A[0]) {
            [C addObject:B[0]];
            [B removeObject:B[0]];
        }
        if (A.count==0) {
            [C addObjectsFromArray:B];
            NSLog(@"C = %@",C);
            index = i+1;
            NSLog(@"index = %d",index);
            return;
        }
        else if (B.count==0) {
            [C addObjectsFromArray:A];
            NSLog(@"C = %@",C);
            index = i+1;
            NSLog(@"index = %d",index);
            return;
        }
    }
    //(2).
    //时间复杂度
    //T(n) = O(f(n)):用"T(n)"表示，"O"为数学符号，f(n)为同数量级，一般是算法中频度最大的语句频度。
    //时间复杂度:T(n) = O(index);
}